Lambert published several examples of continued fractions in this form in 1768, and both Euler and Lagrange investigated similar constructions, but it was Carl Friedrich Gauss who utilized the algebra described in the next section to deduce the general form of this continued fraction, in 1813.
Although Gauss gave the form of this continued fraction, he did not give a proof of its convergence properties. Bernhard Riemann and L.W. Thomé obtained partial results, but the final word on the region in which this continued fraction converges was not given until 1901, by Edward Burr Van Vleck.
Let                               f                      0                          ,                  f                      1                          ,                  f                      2                          ,        …                 be a sequence of analytic functions so that
                              f                      i            −            1                          −                  f                      i                          =                  k                      i                                  z                          f                      i            +            1                                  for all                     i        >        0                , where each                               k                      i                                   is a constant.
Then
                                                        f                              i                −                1                                                    f                              i                                                    =        1        +                  k                      i                          z                                            f                              i                +                1                                                    f                              i                                                    ,                         and so 
                                                        f                              i                                                    f                              i                −                1                                                    =                              1                          1              +                              k                                  i                                            z                                                                    f                                          i                      +                      1                                                                            f                                          i                                                                                                                  Setting                               g                      i                          =                  f                      i                                    /                          f                      i            −            1                                  ,
                              g                      i                          =                              1                          1              +                              k                                  i                                            z                              g                                  i                  +                  1                                                                            ,
So
                              g                      1                          =                                            f                              1                                                    f                              0                                                    =                                                                                                                          1                                                                                                                                                  1                  +                                      k                                          1                                                        z                                      g                                          2                                                                                                          =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                                                  k                                                              1                                                                                      z                                                                                                                                                                                                                                                      1                            +                                                          k                                                              2                                                                                      z                                                          g                                                              3                                                                                                                                                                                                                                            =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                                                  k                                                              1                                                                                      z                                                                                                                                                                                                                                                      1                            +                                                                                                                                                                                                                                                                                                                                                      k                                                                                  2                                                                                                                    z                                                                                                                                                                                                                                                                                                                                                          1                                      +                                                                              k                                                                                  3                                                                                                                    z                                                                              g                                                                                  4                                                                                                                                                                                                                                                                                                                                                                                                                                =        …                         .
Repeating this ad infinitum produces the continued fraction expression
                                                        f                              1                                                    f                              0                                                    =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                                                  k                                                              1                                                                                      z                                                                                                                                                                                                                                                      1                            +                                                                                                                                                                                                                                                                                                                                                      k                                                                                  2                                                                                                                    z                                                                                                                                                                                                                                                                                                                                                          1                                      +                                                                                                                                                                                                                                                                                                                                                                                                                                                          k                                                                                                      3                                                                                                                                                  z                                                                                                                                                                                                                                                                                                                                                                                                                                                              1                                                +                                                                                                                                                ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            In Gauss's continued fraction, the functions                               f                      i                                   are hypergeometric functions of the form                                                               0                                    F                      1                                  ,                                                               1                                    F                      1                                  , and                                                               2                                    F                      1                                  , and the equations                               f                      i            −            1                          −                  f                      i                          =                  k                      i                          z                  f                      i            +            1                                   arise as identities between functions where the parameters differ by integer amounts. These identities can be proven in several ways, for example by expanding out the series and comparing coefficients, or by taking the derivative in several ways and eliminating it from the equations generated.
The simplest case involves
                                                    0                                    F                      1                          (        a        ;        z        )        =        1        +                              1                          a                            1              !                                      z        +                              1                          a              (              a              +              1              )                            2              !                                                z                      2                          +                              1                          a              (              a              +              1              )              (              a              +              2              )                            3              !                                                z                      3                          +        ⋯                         .
Starting with the identity
                                                    0                                    F                      1                          (        a        −        1        ;        z        )        −                                        0                                    F                      1                          (        a        ;        z        )        =                              z                          a              (              a              −              1              )                                                                      0                                    F                      1                          (        a        +        1        ;        z        )                ,
we may take
                              f                      i                          =                                                  0                                    F                      1                          (        a        +        i        ;        z        )        ,                          k                      i                          =                                            1                              (                a                +                i                )                (                a                +                i                −                1                )                                                            ,
giving
                                                                                                          0                                                            F                                  1                                            (              a              +              1              ;              z              )                                                                                        0                                                            F                                  1                                            (              a              ;              z              )                                      =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                                                                                  1                                                                  a                                  (                                  a                                  +                                  1                                  )                                                                                                                      z                                                                                                                                                                                                                                                      1                            +                                                                                                                                                                                                                                                                                                                                                                                                1                                                                                      (                                            a                                            +                                            1                                            )                                            (                                            a                                            +                                            2                                            )                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                          1                                      +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                              1                                                                                                          (                                                      a                                                      +                                                      2                                                      )                                                      (                                                      a                                                      +                                                      3                                                      )                                                                                                                                                                                                      z                                                                                                                                                                                                                                                                                                                                                                                                                                                              1                                                +                                                                                                                                                ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            or
                                                                                                          0                                                            F                                  1                                            (              a              +              1              ;              z              )                                      a                                                                0                                                            F                                  1                                            (              a              ;              z              )                                      =                                                                                                                          1                                                                                                                                                  a                  +                                                                                                                                                                                                                    z                                                                                                                                                                                                                                                      (                            a                            +                            1                            )                            +                                                                                                                                                                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                          (                                      a                                      +                                      2                                      )                                      +                                                                                                                                                                                                                                                                                                                                                                                                        z                                                                                                                                                                                                                                                                                                                                                                                                                                                              (                                                a                                                +                                                3                                                )                                                +                                                                                                                                                ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            .
This expansion converges to the meromorphic function defined by the ratio of the two convergent series (provided, of course, that a is neither zero nor a negative integer).
The next case involves
                                                              1                                    F                      1                          (        a        ;        b        ;        z        )        =        1        +                              a                          b                            1              !                                      z        +                                            a              (              a              +              1              )                                      b              (              b              +              1              )                            2              !                                                z                      2                          +                                            a              (              a              +              1              )              (              a              +              2              )                                      b              (              b              +              1              )              (              b              +              2              )                            3              !                                                z                      3                          +        …                for which the two identities
                                                    1                                    F                      1                          (        a        ;        b        −        1        ;        z        )        −                                        1                                    F                      1                          (        a        +        1        ;        b        ;        z        )        =                                            (              a              −              b              +              1              )              z                                      b              (              b              −              1              )                                                                      1                                    F                      1                          (        a        +        1        ;        b        +        1        ;        z        )                                                                    1                                    F                      1                          (        a        ;        b        −        1        ;        z        )        −                                        1                                    F                      1                          (        a        ;        b        ;        z        )        =                                            a              z                                      b              (              b              −              1              )                                                                      1                                    F                      1                          (        a        +        1        ;        b        +        1        ;        z        )                are used alternately.
Let
                              f                      0                          (        z        )        =                                        1                                    F                      1                          (        a        ;        b        ;        z        )                ,
                              f                      1                          (        z        )        =                                        1                                    F                      1                          (        a        +        1        ;        b        +        1        ;        z        )                ,
                              f                      2                          (        z        )        =                                        1                                    F                      1                          (        a        +        1        ;        b        +        2        ;        z        )                ,
                              f                      3                          (        z        )        =                                        1                                    F                      1                          (        a        +        2        ;        b        +        3        ;        z        )                ,
                              f                      4                          (        z        )        =                                        1                                    F                      1                          (        a        +        2        ;        b        +        4        ;        z        )                ,
etc.
This gives                               f                      i            −            1                          −                  f                      i                          =                  k                      i                          z                  f                      i            +            1                                   where                               k                      1                          =                                                            a                −                b                                            b                (                b                +                1                )                                                    ,                  k                      2                          =                                                            a                +                1                                            (                b                +                1                )                (                b                +                2                )                                                    ,                  k                      3                          =                                                            a                −                b                −                1                                            (                b                +                2                )                (                b                +                3                )                                                    ,                  k                      4                          =                                                            a                +                2                                            (                b                +                3                )                (                b                +                4                )                                                            , producing
                                                                                                                          1                                                            F                                  1                                            (              a              +              1              ;              b              +              1              ;              z              )                                                                                                        1                                                            F                                  1                                            (              a              ;              b              ;              z              )                                      =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                                                                                                                    a                                  −                                  b                                                                                                  b                                  (                                  b                                  +                                  1                                  )                                                                                                                      z                                                                                                                                                                                                                                                      1                            +                                                                                                                                                                                                                                                                                                                                                                                                                                            a                                            +                                            1                                                                                                                                (                                            b                                            +                                            1                                            )                                            (                                            b                                            +                                            2                                            )                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                          1                                      +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    a                                                      −                                                      b                                                      −                                                      1                                                                                                                                                              (                                                      b                                                      +                                                      2                                                      )                                                      (                                                      b                                                      +                                                      3                                                      )                                                                                                                                                                                                      z                                                                                                                                                                                                                                                                                                                                                                                                                                                              1                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            a                                                                +                                                                2                                                                                                                                                                                            (                                                                b                                                                +                                                                3                                                                )                                                                (                                                                b                                                                +                                                                4                                                                )                                                                                                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  1                                                          +                                                                                                                                                                              ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                or
                                                                                                                          1                                                            F                                  1                                            (              a              +              1              ;              b              +              1              ;              z              )                                      b                                                                                1                                                            F                                  1                                            (              a              ;              b              ;              z              )                                      =                                                                                                                          1                                                                                                                                                  b                  +                                                                                                                                                                                                                    (                            a                            −                            b                            )                            z                                                                                                                                                                                                                                                      (                            b                            +                            1                            )                            +                                                                                                                                                                                                                                                                                                              (                                      a                                      +                                      1                                      )                                      z                                                                                                                                                                                                                                                                                                                                                          (                                      b                                      +                                      2                                      )                                      +                                                                                                                                                                                                                                                                                                                                                                                                        (                                                a                                                −                                                b                                                −                                                1                                                )                                                z                                                                                                                                                                                                                                                                                                                                                                                                                                                              (                                                b                                                +                                                3                                                )                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  (                                                          a                                                          +                                                          2                                                          )                                                          z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  (                                                          b                                                          +                                                          4                                                          )                                                          +                                                                                                                                                                              ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                Similarly
                                                                                                                          1                                                            F                                  1                                            (              a              ;              b              +              1              ;              z              )                                                                                                        1                                                            F                                  1                                            (              a              ;              b              ;              z              )                                      =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                                                                                  a                                                                  b                                  (                                  b                                  +                                  1                                  )                                                                                                                      z                                                                                                                                                                                                                                                      1                            +                                                                                                                                                                                                                                                                                                                                                                                                                                            a                                            −                                            b                                            −                                            1                                                                                                                                (                                            b                                            +                                            1                                            )                                            (                                            b                                            +                                            2                                            )                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                          1                                      +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    a                                                      +                                                      1                                                                                                                                                              (                                                      b                                                      +                                                      2                                                      )                                                      (                                                      b                                                      +                                                      3                                                      )                                                                                                                                                                                                      z                                                                                                                                                                                                                                                                                                                                                                                                                                                              1                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            a                                                                −                                                                b                                                                −                                                                2                                                                                                                                                                                            (                                                                b                                                                +                                                                3                                                                )                                                                (                                                                b                                                                +                                                                4                                                                )                                                                                                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  1                                                          +                                                                                                                                                                              ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                or
                                                                                                                          1                                                            F                                  1                                            (              a              ;              b              +              1              ;              z              )                                      b                                                                                1                                                            F                                  1                                            (              a              ;              b              ;              z              )                                      =                                                                                                                          1                                                                                                                                                  b                  +                                                                                                                                                                                                                    a                            z                                                                                                                                                                                                                                                      (                            b                            +                            1                            )                            +                                                                                                                                                                                                                                                                                                              (                                      a                                      −                                      b                                      −                                      1                                      )                                      z                                                                                                                                                                                                                                                                                                                                                          (                                      b                                      +                                      2                                      )                                      +                                                                                                                                                                                                                                                                                                                                                                                                        (                                                a                                                +                                                1                                                )                                                z                                                                                                                                                                                                                                                                                                                                                                                                                                                              (                                                b                                                +                                                3                                                )                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  (                                                          a                                                          −                                                          b                                                          −                                                          2                                                          )                                                          z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  (                                                          b                                                          +                                                          4                                                          )                                                          +                                                                                                                                                                              ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                Since                                                               1                                    F                      1                          (        0        ;        b        ;        z        )        =        1                , setting a to 0 and replacing b + 1 with b in the first continued fraction gives a simplified special case:
                                                              1                                    F                      1                          (        1        ;        b        ;        z        )        =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                    −                            z                                                                                                                                                                                                                                                      b                            +                                                                                                                                                                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                          (                                      b                                      +                                      1                                      )                                      +                                                                                                                                                                                                                                                                                                                                                                                                        −                                                b                                                z                                                                                                                                                                                                                                                                                                                                                                                                                                                              (                                                b                                                +                                                2                                                )                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  2                                                          z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  (                                                          b                                                          +                                                          3                                                          )                                                          +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            −                                                                    (                                                                    b                                                                    +                                                                    1                                                                    )                                                                    z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      (                                                                    b                                                                    +                                                                    4                                                                    )                                                                    +                                                                                                                                                                                                            ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      The final case involves
                                                              2                                    F                      1                          (        a        ,        b        ;        c        ;        z        )        =        1        +                                            a              b                                      c                            1              !                                      z        +                                            a              (              a              +              1              )              b              (              b              +              1              )                                      c              (              c              +              1              )                            2              !                                                z                      2                          +                                            a              (              a              +              1              )              (              a              +              2              )              b              (              b              +              1              )              (              b              +              2              )                                      c              (              c              +              1              )              (              c              +              2              )                            3              !                                                z                      3                          +        …                        .
Again, two identities are used alternately.
                                                    2                                    F                      1                          (        a        ,        b        ;        c        −        1        ;        z        )        −                                        2                                    F                      1                          (        a        +        1        ,        b        ;        c        ;        z        )        =                                            (              a              −              c              +              1              )              b              z                                      c              (              c              −              1              )                                                                      2                                    F                      1                          (        a        +        1        ,        b        +        1        ;        c        +        1        ;        z        )                ,
                                                    2                                    F                      1                          (        a        ,        b        ;        c        −        1        ;        z        )        −                                        2                                    F                      1                          (        a        ,        b        +        1        ;        c        ;        z        )        =                                            (              b              −              c              +              1              )              a              z                                      c              (              c              −              1              )                                                                      2                                    F                      1                          (        a        +        1        ,        b        +        1        ;        c        +        1        ;        z        )                .
These are essentially the same identity with a and b interchanged.
Let
                              f                      0                          (        z        )        =                                        2                                    F                      1                          (        a        ,        b        ;        c        ;        z        )                ,
                              f                      1                          (        z        )        =                                        2                                    F                      1                          (        a        +        1        ,        b        ;        c        +        1        ;        z        )                ,
                              f                      2                          (        z        )        =                                        2                                    F                      1                          (        a        +        1        ,        b        +        1        ;        c        +        2        ;        z        )                ,
                              f                      3                          (        z        )        =                                        2                                    F                      1                          (        a        +        2        ,        b        +        1        ;        c        +        3        ;        z        )                ,
                              f                      4                          (        z        )        =                                        2                                    F                      1                          (        a        +        2        ,        b        +        2        ;        c        +        4        ;        z        )                ,
etc.
This gives                               f                      i            −            1                          −                  f                      i                          =                  k                      i                          z                  f                      i            +            1                                   where                               k                      1                          =                                                            (                a                −                c                )                b                                            c                (                c                +                1                )                                                    ,                  k                      2                          =                                                            (                b                −                c                −                1                )                (                a                +                1                )                                            (                c                +                1                )                (                c                +                2                )                                                    ,                  k                      3                          =                                                            (                a                −                c                −                1                )                (                b                +                1                )                                            (                c                +                2                )                (                c                +                3                )                                                    ,                  k                      4                          =                                                            (                b                −                c                −                2                )                (                a                +                2                )                                            (                c                +                3                )                (                c                +                4                )                                                            , producing
                                                                                                                          2                                                            F                                  1                                            (              a              +              1              ,              b              ;              c              +              1              ;              z              )                                                                                                        2                                                            F                                  1                                            (              a              ,              b              ;              c              ;              z              )                                      =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                                                                                                                    (                                  a                                  −                                  c                                  )                                  b                                                                                                  c                                  (                                  c                                  +                                  1                                  )                                                                                                                      z                                                                                                                                                                                                                                                      1                            +                                                                                                                                                                                                                                                                                                                                                                                                                                            (                                            b                                            −                                            c                                            −                                            1                                            )                                            (                                            a                                            +                                            1                                            )                                                                                                                                (                                            c                                            +                                            1                                            )                                            (                                            c                                            +                                            2                                            )                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                          1                                      +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    (                                                      a                                                      −                                                      c                                                      −                                                      1                                                      )                                                      (                                                      b                                                      +                                                      1                                                      )                                                                                                                                                              (                                                      c                                                      +                                                      2                                                      )                                                      (                                                      c                                                      +                                                      3                                                      )                                                                                                                                                                                                      z                                                                                                                                                                                                                                                                                                                                                                                                                                                              1                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            (                                                                b                                                                −                                                                c                                                                −                                                                2                                                                )                                                                (                                                                a                                                                +                                                                2                                                                )                                                                                                                                                                                            (                                                                c                                                                +                                                                3                                                                )                                                                (                                                                c                                                                +                                                                4                                                                )                                                                                                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  1                                                          +                                                                                                                                                                              ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                or
                                                                                                                          2                                                            F                                  1                                            (              a              +              1              ,              b              ;              c              +              1              ;              z              )                                      c                                                                                2                                                            F                                  1                                            (              a              ,              b              ;              c              ;              z              )                                      =                                                                                                                          1                                                                                                                                                  c                  +                                                                                                                                                                                                                    (                            a                            −                            c                            )                            b                            z                                                                                                                                                                                                                                                      (                            c                            +                            1                            )                            +                                                                                                                                                                                                                                                                                                              (                                      b                                      −                                      c                                      −                                      1                                      )                                      (                                      a                                      +                                      1                                      )                                      z                                                                                                                                                                                                                                                                                                                                                          (                                      c                                      +                                      2                                      )                                      +                                                                                                                                                                                                                                                                                                                                                                                                        (                                                a                                                −                                                c                                                −                                                1                                                )                                                (                                                b                                                +                                                1                                                )                                                z                                                                                                                                                                                                                                                                                                                                                                                                                                                              (                                                c                                                +                                                3                                                )                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  (                                                          b                                                          −                                                          c                                                          −                                                          2                                                          )                                                          (                                                          a                                                          +                                                          2                                                          )                                                          z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  (                                                          c                                                          +                                                          4                                                          )                                                          +                                                                                                                                                                              ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                Since                                                               2                                    F                      1                          (        0        ,        b        ;        c        ;        z        )        =        1                , setting a to 0 and replacing c + 1 with c gives a simplified special case of the continued fraction:
                                                              2                                    F                      1                          (        1        ,        b        ;        c        ;        z        )        =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                    −                            b                            z                                                                                                                                                                                                                                                      c                            +                                                                                                                                                                                                                                                                                                              (                                      b                                      −                                      c                                      )                                      z                                                                                                                                                                                                                                                                                                                                                          (                                      c                                      +                                      1                                      )                                      +                                                                                                                                                                                                                                                                                                                                                                                                        −                                                c                                                (                                                b                                                +                                                1                                                )                                                z                                                                                                                                                                                                                                                                                                                                                                                                                                                              (                                                c                                                +                                                2                                                )                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  2                                                          (                                                          b                                                          −                                                          c                                                          −                                                          1                                                          )                                                          z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  (                                                          c                                                          +                                                          3                                                          )                                                          +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            −                                                                    (                                                                    c                                                                    +                                                                    1                                                                    )                                                                    (                                                                    b                                                                    +                                                                    2                                                                    )                                                                    z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      (                                                                    c                                                                    +                                                                    4                                                                    )                                                                    +                                                                                                                                                                                                            ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      In this section, the cases where one or more of the parameters is a negative integer are excluded, since in these cases either the hypergeometric series are undefined or that they are polynomials so the continued fraction terminates. Other trivial exceptions are excluded as well.
In the cases                                                               0                                    F                      1                                   and                                                               1                                    F                      1                                  , the series converge everywhere so the fraction on the left hand side is a meromorphic function. The continued fractions on the right hand side will converge uniformly on any closed and bounded set that contains no poles of this function.
In the case                                                               2                                    F                      1                                  , the radius of convergence of the series is 1 and the fraction on the left hand side is a meromorphic function within this circle. The continued fractions on the right hand side will converge to the function everywhere inside this circle.
Outside the circle, the continued fraction represents the analytic continuation of the function to the complex plane with the positive real axis, from +1 to the point at infinity removed. In most cases +1 is a branch point and the line from +1 to positive infinity is a branch cut for this function. The continued fraction converges to a meromorphic function on this domain, and it converges uniformly on any closed and bounded subset of this domain that does not contain any poles.
We have
                    cosh                (        z        )        =                                        0                                    F                      1                          (                                            1              2                                      ;                                                            z                                  2                                            4                                      )        ,                                    sinh                (        z        )        =        z                                        0                                    F                      1                          (                                            3              2                                      ;                                                            z                                  2                                            4                                      )        ,                so
                    tanh                (        z        )        =                                            z                                                                0                                                            F                                  1                                            (                                                                    3                    2                                                              ;                                                                                          z                                              2                                                              4                                                              )                                                                                        0                                                            F                                  1                                            (                                                                    1                    2                                                              ;                                                                                          z                                              2                                                              4                                                              )                                      =                                                                                                                          z                                      /                                    2                                                                                                                                                                                                                    1                        2                                                                              +                                                                                                                                                                                                                                                                                                                    z                                                                      2                                                                                                  4                                                                                                                                                                                                                                                                                                                                                                                                                3                                  2                                                                                                                      +                                                                                                                                                                                                                                                                                                                                                                                                                                            z                                                                                          2                                                                                                                                4                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      5                                            2                                                                                                                                                              +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    z                                                                                                              2                                                                                                                                                              4                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            7                                                      2                                                                                                                                                                                                      +                                                                                                                                                ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    =                                                                                                                          z                                                                                                                                                  1                  +                                                                                                                                                                                                                                                  z                                                              2                                                                                                                                                                                                                                                                                                                3                            +                                                                                                                                                                                                                                                                                                                                                      z                                                                                  2                                                                                                                                                                                                                                                                                                                                                                                                                                        5                                      +                                                                                                                                                                                                                                                                                                                                                                                                                                                          z                                                                                                      2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                7                                                +                                                                                                                                                ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    .                This particular expansion is known as Lambert's continued fraction and dates back to 1768.
It easily follows that
                    tan                (        z        )        =                                                                                                                          z                                                                                                                                                  1                  −                                                                                                                                                                                                                                                  z                                                              2                                                                                                                                                                                                                                                                                                                3                            −                                                                                                                                                                                                                                                                                                                                                      z                                                                                  2                                                                                                                                                                                                                                                                                                                                                                                                                                        5                                      −                                                                                                                                                                                                                                                                                                                                                                                                                                                          z                                                                                                      2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                7                                                −                                                                                                                                                ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    .                The expansion of tanh can be used to prove that en is irrational for every integer n (which is alas not enough to prove that e is transcendental). The expansion of tan was used by both Lambert and Legendre to prove that π is irrational.
The Bessel function                               J                      ν                                   can be written
                              J                      ν                          (        z        )        =                                            (                                                                    1                    2                                                              z                              )                                  ν                                                                    Γ              (              ν              +              1              )                                                                      0                                    F                      1                          (        ν        +        1        ;        −                                            z                              2                                      4                          )        ,                from which it follows
                                                                        J                                  ν                                            (              z              )                                                      J                                  ν                  −                  1                                            (              z              )                                      =                                                                                                                          z                                                                                                                                                  2                  ν                  −                                                                                                                                                                                                                                                  z                                                              2                                                                                                                                                                                                                                                                                                                2                            (                            ν                            +                            1                            )                            −                                                                                                                                                                                                                                                                                                                                                      z                                                                                  2                                                                                                                                                                                                                                                                                                                                                                                                                                        2                                      (                                      ν                                      +                                      2                                      )                                      −                                                                                                                                                                                                                                                                                                                                                                                                                                                          z                                                                                                      2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                2                                                (                                                ν                                                +                                                3                                                )                                                −                                                                                                                                                ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    .                These formulas are also valid for every complex z.
Since                               e                      z                          =                                                  1                                    F                      1                          (        1        ;        1        ;        z        )                ,                     1                  /                          e                      z                          =                  e                      −            z                                  
                              e                      z                          =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                    −                            z                                                                                                                                                                                                                                                      1                            +                                                                                                                                                                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                          2                                      +                                                                                                                                                                                                                                                                                                                                                                                                        −                                                z                                                                                                                                                                                                                                                                                                                                                                                                                                                              3                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  2                                                          z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  4                                                          +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            −                                                                    2                                                                    z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      5                                                                    +                                                                                                                                                                                                            ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    e                      z                          =        1        +                                                                                                                          z                                                                                                                                                  1                  +                                                                                                                                                                                                                    −                            z                                                                                                                                                                                                                                                      2                            +                                                                                                                                                                                                                                                                                                              z                                                                                                                                                                                                                                                                                                                                                          3                                      +                                                                                                                                                                                                                                                                                                                                                                                                        −                                                2                                                z                                                                                                                                                                                                                                                                                                                                                                                                                                                              4                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  2                                                          z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  5                                                          +                                                                                                                                                                              ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                .
With some manipulation, this can be used to prove the simple continued fraction representation of e,
                    e        =        2        +                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                    1                                                                                                                                                                                                                                                      2                            +                                                                                                                                                                                                                                                                                                              1                                                                                                                                                                                                                                                                                                                                                          1                                      +                                                                                                                                                                                                                                                                                                                                                                                                        1                                                                                                                                                                                                                                                                                                                                                                                                                                                              1                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  1                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  4                                                          +                                                                                                                                                                              ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                The error function erf (z), given by
                    erf                (        z        )        =                              2                          π                                                ∫                      0                                z                                    e                      −                          t                              2                                                    d        t        ,                can also be computed in terms of Kummer's hypergeometric function:
                    erf                (        z        )        =                                            2              z                                      π                                                e                      −                          z                              2                                                                                    1                                    F                      1                          (        1        ;                                                            3                2                                                    ;                  z                      2                          )        .                By applying the continued fraction of Gauss, a useful expansion valid for every complex number z can be obtained:
                                                        π                        2                                    e                                    z                              2                                                    erf                (        z        )        =                                                                                                                          z                                                                                                                                                  1                  −                                                                                                                                                                                                                                                  z                                                              2                                                                                                                                                                                                                                                                                                                                                                              3                                2                                                                                      +                                                                                                                                                                                                                                                                                                                                                      z                                                                                  2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                          5                                          2                                                                                                                    −                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                              3                                                    2                                                                                                                                                                                                    z                                                                                                      2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      7                                                    2                                                                                                                                                  +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  2                                                                                                                      z                                                                                                                          2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  9                                                              2                                                                                                                                                                                −                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                          5                                                                        2                                                                                                                                                                                                                                                                                    z                                                                                                                                              2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                              11                                                                        2                                                                                                                                                                                                              +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      3                                                                                                                                                              z                                                                                                                                                                  2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                          13                                                                                  2                                                                                                                                                                                                                                            −                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      7                                                                                            2                                                                                                                                                                                                                                                                                                                                                                    z                                                                                                                                                                                      2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      15                                                                                            2                                                                                                                                                                                                                                                                          +                                                                                        −                                                                                        ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                .                A similar argument can be made to derive continued fraction expansions for the Fresnel integrals, for the Dawson function, and for the incomplete gamma function. A simpler version of the argument yields two useful continued fraction expansions of the exponential function.
From
                    (        1        −        z                  )                      −            b                          =                                                  1                                    F                      0                          (        b        ;        ;        z        )        =                                        2                                    F                      1                          (        1        ,        b        ;        1        ;        z        )                ,
                    (        1        −        z                  )                      −            b                          =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                    −                            b                            z                                                                                                                                                                                                                                                      1                            +                                                                                                                                                                                                                                                                                                              (                                      b                                      −                                      1                                      )                                      z                                                                                                                                                                                                                                                                                                                                                          2                                      +                                                                                                                                                                                                                                                                                                                                                                                                        −                                                (                                                b                                                +                                                1                                                )                                                z                                                                                                                                                                                                                                                                                                                                                                                                                                                              3                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  2                                                          (                                                          b                                                          −                                                          2                                                          )                                                          z                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  4                                                          +                                                                                                                                                                              ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                It is easily shown that the Taylor series expansion of arctan z in a neighborhood of zero is given by
                    arctan                z        =        z        F        (                                                            1                2                                                    ,        1        ;                                                            3                2                                                    ;        −                  z                      2                          )        .                The continued fraction of Gauss can be applied to this identity, yielding the expansion
                    arctan                z        =                                                                                                                          z                                                                                                                                                  1                  +                                                                                                                                                                                                                    (                            1                            z                                                          )                                                              2                                                                                                                                                                                                                                                                                                                3                            +                                                                                                                                                                                                                                                                                                              (                                      2                                      z                                                                              )                                                                                  2                                                                                                                                                                                                                                                                                                                                                                                                                                        5                                      +                                                                                                                                                                                                                                                                                                                                                                                                        (                                                3                                                z                                                                                                  )                                                                                                      2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                7                                                +                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  (                                                          4                                                          z                                                                                                                      )                                                                                                                          2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        9                                                          +                                                          ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        ,                which converges to the principal branch of the inverse tangent function on the cut complex plane, with the cut extending along the imaginary axis from i to the point at infinity, and from −i to the point at infinity.
This particular continued fraction converges fairly quickly when z = 1, giving the value π/4 to seven decimal places by the ninth convergent. The corresponding series
                                          π            4                          =                                                                                                                          1                                                                                                                                                  1                  +                                                                                                                                                                                                                                                  1                                                              2                                                                                                                                                                                                                                                                                                                2                            +                                                                                                                                                                                                                                                                                                                                                      3                                                                                  2                                                                                                                                                                                                                                                                                                                                                                                                                                        2                                      +                                                                                                                                                                                                                                                                                                                                                                                                                                                          5                                                                                                      2                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                2                                                +                                                ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    =        1        −                              1            3                          +                              1            5                          −                              1            7                          +        −        …                converges much more slowly, with more than a million terms needed to yield seven decimal places of accuracy.
Variations of this argument can be used to produce continued fraction expansions for the natural logarithm, the arcsin function, and the generalized binomial series.