This list of mathematical series contains formulae for finite and infinite sums. It can be used in conjunction with other tools for evaluating sums.
Here,                               0                      0                                   is taken to have the value                     1                                              B                      n                          (        x        )                 is a Bernoulli polynomial.                              B                      n                                   is a Bernoulli number, and here,                               B                      1                          =        −                              1            2                          .                                              E                      n                                   is an Euler number.                    ζ        (        s        )                 is the Riemann zeta function.                    Γ        (        z        )                 is the gamma function.                              ψ                      n                          (        z        )                 is a polygamma function.                              Li                      s                                  (        z        )                 is a polylogarithm.See Faulhaber's formula.
                              ∑                      k            =            0                                m                                    k                      n            −            1                          =                                                            B                                  n                                            (              m              +              1              )              −                              B                                  n                                                      n                                                  The first few values are:
                              ∑                      k            =            1                                m                          k        =                                            m              (              m              +              1              )                        2                                                                                ∑                      k            =            1                                m                                    k                      2                          =                                            m              (              m              +              1              )              (              2              m              +              1              )                        6                          =                                            m                              3                                      3                          +                                            m                              2                                      2                          +                              m            6                                                                                ∑                      k            =            1                                m                                    k                      3                          =                              [                                                            m                  (                  m                  +                  1                  )                                2                                      ]                                2                          =                                            m                              4                                      4                          +                                            m                              3                                      2                          +                                            m                              2                                      4                                                  See zeta constants.
                    ζ        (        2        n        )        =                  ∑                      k            =            1                                ∞                                                1                          k                              2                n                                                    =        (        −        1                  )                      n            +            1                                                                              B                                  2                  n                                            (              2              π                              )                                  2                  n                                                                    2              (              2              n              )              !                                              The first few values are:
                    ζ        (        2        )        =                  ∑                      k            =            1                                ∞                                                1                          k                              2                                                    =                                            π                              2                                      6                                                   (the Basel problem)                    ζ        (        4        )        =                  ∑                      k            =            1                                ∞                                                1                          k                              4                                                    =                                            π                              4                                      90                                                                      ζ        (        6        )        =                  ∑                      k            =            1                                ∞                                                1                          k                              6                                                    =                                            π                              6                                      945                                                  Finite sums:
                              ∑                      k            =            0                                n                                    z                      k                          =                                            1              −                              z                                  n                  +                  1                                                                    1              −              z                                                              , (geometric series)                              ∑                      k            =            1                                n                          k                  z                      k                          =        z                                            1              −              (              n              +              1              )                              z                                  n                                            +              n                              z                                  n                  +                  1                                                                    (              1              −              z                              )                                  2                                                                                                                          ∑                      k            =            1                                n                                    k                      2                                    z                      k                          =        z                                            1              +              z              −              (              n              +              1                              )                                  2                                                            z                                  n                                            +              (              2                              n                                  2                                            +              2              n              −              1              )                              z                                  n                  +                  1                                            −                              n                                  2                                                            z                                  n                  +                  2                                                                    (              1              −              z                              )                                  3                                                                                                                          ∑                      k            =            1                                n                                    k                      m                                    z                      k                          =                              (            z                                          d                                  d                  z                                                      )                                m                                                              1              −                              z                                  n                  +                  1                                                                    1              −              z                                              Infinite sums, valid for                               |                z                  |                <        1                 (see polylogarithm):
                              Li                      n                                  (        z        )        =                  ∑                      k            =            1                                ∞                                                              z                              k                                                    k                              n                                                                            The following is a useful property to calculate low-integer-order polylogarithms recursively in closed form:
                                          d                          d              z                                                Li                      n                                  (        z        )        =                                                            Li                                  n                  −                  1                                                          (              z              )                        z                                                                                Li                      1                                  (        z        )        =                  ∑                      k            =            1                                ∞                                                              z                              k                                      k                          =        −        ln                (        1        −        z        )                                                      Li                      0                                  (        z        )        =                  ∑                      k            =            1                                ∞                                    z                      k                          =                              z                          1              −              z                                                                                    Li                      −            1                                  (        z        )        =                  ∑                      k            =            1                                ∞                          k                  z                      k                          =                              z                          (              1              −              z                              )                                  2                                                                                                                          Li                      −            2                                  (        z        )        =                  ∑                      k            =            1                                ∞                                    k                      2                                    z                      k                          =                                            z              (              1              +              z              )                                      (              1              −              z                              )                                  3                                                                                                                          Li                      −            3                                  (        z        )        =                  ∑                      k            =            1                                ∞                                    k                      3                                    z                      k                          =                                            z              (              1              +              4              z              +                              z                                  2                                            )                                      (              1              −              z                              )                                  4                                                                                                                          Li                      −            4                                  (        z        )        =                  ∑                      k            =            1                                ∞                                    k                      4                                    z                      k                          =                                            z              (              1              +              z              )              (              1              +              10              z              +                              z                                  2                                            )                                      (              1              −              z                              )                                  5                                                                                                                          ∑                      k            =            0                                ∞                                                              z                              k                                                    k              !                                      =                  e                      z                                                                                ∑                      k            =            0                                ∞                          k                                            z                              k                                                    k              !                                      =        z                  e                      z                                                   (cf. mean of Poisson distribution)                              ∑                      k            =            0                                ∞                                    k                      2                                                              z                              k                                                    k              !                                      =        (        z        +                  z                      2                          )                  e                      z                                                   (cf. second moment of Poisson distribution)                              ∑                      k            =            0                                ∞                                    k                      3                                                              z                              k                                                    k              !                                      =        (        z        +        3                  z                      2                          +                  z                      3                          )                  e                      z                                                                                ∑                      k            =            0                                ∞                                    k                      4                                                              z                              k                                                    k              !                                      =        (        z        +        7                  z                      2                          +        6                  z                      3                          +                  z                      4                          )                  e                      z                                                                                ∑                      k            =            0                                ∞                                    k                      n                                                              z                              k                                                    k              !                                      =        z                              d                          d              z                                                ∑                      k            =            0                                ∞                                    k                      n            −            1                                                              z                              k                                                    k              !                                                      =                  e                      z                                    T                      n                          (        z        )                where                               T                      n                          (        z        )                 is the Touchard polynomials.
Trigonometric, inverse trigonometric, hyperbolic, and inverse hyperbolic functions
                              ∑                      k            =            0                                ∞                                                              (              −              1                              )                                  k                                                            z                                  2                  k                  +                  1                                                                    (              2              k              +              1              )              !                                      =        sin                z                                                              ∑                      k            =            0                                ∞                                                              z                              2                k                +                1                                                    (              2              k              +              1              )              !                                      =        sinh                z                                                              ∑                      k            =            0                                ∞                                                              (              −              1                              )                                  k                                                            z                                  2                  k                                                                    (              2              k              )              !                                      =        cos                z                                                              ∑                      k            =            0                                ∞                                                              z                              2                k                                                    (              2              k              )              !                                      =        cosh                z                                                              ∑                      k            =            1                                ∞                                                              (              −              1                              )                                  k                  −                  1                                            (                              2                                  2                  k                                            −              1              )                              2                                  2                  k                                                            B                                  2                  k                                                            z                                  2                  k                  −                  1                                                                    (              2              k              )              !                                      =        tan                z        ,                  |                z                  |                <                              π            2                                                                                ∑                      k            =            1                                ∞                                                              (                              2                                  2                  k                                            −              1              )                              2                                  2                  k                                                            B                                  2                  k                                                            z                                  2                  k                  −                  1                                                                    (              2              k              )              !                                      =        tanh                z        ,                  |                z                  |                <                              π            2                                                                                ∑                      k            =            0                                ∞                                                              (              −              1                              )                                  k                                                            2                                  2                  k                                                            B                                  2                  k                                                            z                                  2                  k                  −                  1                                                                    (              2              k              )              !                                      =        cot                z        ,                  |                z                  |                <        π                                                              ∑                      k            =            0                                ∞                                                                              2                                  2                  k                                                            B                                  2                  k                                                            z                                  2                  k                  −                  1                                                                    (              2              k              )              !                                      =        coth                z        ,                  |                z                  |                <        π                                                              ∑                      k            =            0                                ∞                                                              (              −              1                              )                                  k                  −                  1                                            (                              2                                  2                  k                                            −              2              )                              B                                  2                  k                                                            z                                  2                  k                  −                  1                                                                    (              2              k              )              !                                      =        csc                z        ,                  |                z                  |                <        π                                                              ∑                      k            =            0                                ∞                                                              −              (                              2                                  2                  k                                            −              2              )                              B                                  2                  k                                                            z                                  2                  k                  −                  1                                                                    (              2              k              )              !                                      =        csch                z        ,                  |                z                  |                <        π                                                              ∑                      k            =            0                                ∞                                                              (              −              1                              )                                  k                                                            E                                  2                  k                                                            z                                  2                  k                                                                    (              2              k              )              !                                      =        sec                z        ,                  |                z                  |                <                              π            2                                                                                ∑                      k            =            0                                ∞                                                                              E                                  2                  k                                                            z                                  2                  k                                                                    (              2              k              )              !                                      =        sech                z        ,                  |                z                  |                <                              π            2                                                                                ∑                      k            =            1                                ∞                                                              (              −              1                              )                                  k                  −                  1                                                            z                                  2                  k                                                                    (              2              k              )              !                                      =        ver                z                                 (versine)                              ∑                      k            =            1                                ∞                                                              (              −              1                              )                                  k                  −                  1                                                            z                                  2                  k                                                                    2              (              2              k              )              !                                      =        hav                z                                 (haversine)                              ∑                      k            =            0                                ∞                                                              (              2              k              )              !                              z                                  2                  k                  +                  1                                                                                    2                                  2                  k                                            (              k              !                              )                                  2                                            (              2              k              +              1              )                                      =        arcsin                z        ,                  |                z                  |                ≤        1                                                              ∑                      k            =            0                                ∞                                                              (              −              1                              )                                  k                                            (              2              k              )              !                              z                                  2                  k                  +                  1                                                                                    2                                  2                  k                                            (              k              !                              )                                  2                                            (              2              k              +              1              )                                      =        arcsinh                          z                ,                  |                z                  |                ≤        1                                                              ∑                      k            =            0                                ∞                                                              (              −              1                              )                                  k                                                            z                                  2                  k                  +                  1                                                                    2              k              +              1                                      =        arctan                z        ,                  |                z                  |                <        1                                                              ∑                      k            =            0                                ∞                                                              z                              2                k                +                1                                                    2              k              +              1                                      =        arctanh                z        ,                  |                z                  |                <        1                                                    ln                2        +                  ∑                      k            =            1                                ∞                                                              (              −              1                              )                                  k                  −                  1                                            (              2              k              )              !                              z                                  2                  k                                                                                    2                                  2                  k                  +                  1                                            k              (              k              !                              )                                  2                                                                    =        ln                          (          1          +                                    1              +                              z                                  2                                                              )                ,                  |                z                  |                ≤        1                                                              ∑                      k            =            0                                ∞                                                              (              4              k              )              !                                                      2                                  4                  k                                                                              2                                            (              2              k              )              !              (              2              k              +              1              )              !                                                z                      k                          =                                                            1                −                                                      1                    −                    z                                                              z                                      ,                  |                z                  |                <        1                                              ∑                      k            =            0                                ∞                                                                              2                                  2                  k                                            (              k              !                              )                                  2                                                                    (              k              +              1              )              (              2              k              +              1              )              !                                                z                      2            k            +            2                          =                              (            arcsin                                      z                        )                                2                          ,                  |                z                  |                ≤        1                                              ∑                      n            =            0                                ∞                                                                              ∏                                  k                  =                  0                                                  n                  −                  1                                            (              4                              k                                  2                                            +                              α                                  2                                            )                                      (              2              n              )              !                                                z                      2            n                          +                  ∑                      n            =            0                                ∞                                                              α                              ∏                                  k                  =                  0                                                  n                  −                  1                                            [              (              2              k              +              1                              )                                  2                                            +                              α                                  2                                            ]                                      (              2              n              +              1              )              !                                                z                      2            n            +            1                          =                  e                      α            arcsin                                      z                                      ,                  |                z                  |                ≤        1                                    (        1        +        z                  )                      α                          =                  ∑                      k            =            0                                ∞                                                              (                                      α              k                                      )                                                z                      k                          ,                  |                z                  |                <        1                 (see Binomial theorem)                               ∑                      k            =            0                                ∞                                                              (                                                      α                +                k                −                1                            k                                      )                                                z                      k                          =                              1                          (              1              −              z                              )                                  α                                                                    ,                  |                z                  |                <        1                                               ∑                      k            =            0                                ∞                                                1                          k              +              1                                                                          (                                                      2                k                            k                                      )                                                z                      k                          =                                            1              −                                                1                  −                  4                  z                                                                    2              z                                      ,                  |                z                  |                ≤                              1            4                                  , generating function of the Catalan numbers                               ∑                      k            =            0                                ∞                                                              (                                                      2                k                            k                                      )                                                z                      k                          =                              1                          1              −              4              z                                      ,                  |                z                  |                <                              1            4                                  , generating function of the Central binomial coefficients                               ∑                      k            =            0                                ∞                                                              (                                                      2                k                +                α                            k                                      )                                                z                      k                          =                              1                          1              −              4              z                                                            (                                                            1                  −                                                            1                      −                      4                      z                                                                                        2                  z                                                      )                                α                          ,                  |                z                  |                <                              1            4                                                                ∑                      k            =            1                                ∞                                    H                      k                                    z                      k                          =                                            −              ln                            (              1              −              z              )                                      1              −              z                                      ,                  |                z                  |                <        1                                              ∑                      k            =            1                                ∞                                                              H                              k                                                    k              +              1                                                z                      k            +            1                          =                              1            2                                                [            ln                        (            1            −            z            )            ]                                2                          ,                          |                z                  |                <        1                                              ∑                      k            =            1                                ∞                                                              (              −              1                              )                                  k                  −                  1                                                            H                                  2                  k                                                                    2              k              +              1                                                z                      2            k            +            1                          =                              1            2                          arctan                          z                log                          (          1          +                      z                          2                                )                ,                          |                z                  |                <        1                                              ∑                      n            =            0                                ∞                                    ∑                      k            =            0                                2            n                                                              (              −              1                              )                                  k                                                                    2              k              +              1                                                                          z                              4                n                +                2                                                    4              n              +              2                                      =                              1            4                          arctan                          z                log                                                    1              +              z                                      1              −              z                                      ,                          |                z                  |                <        1                                              ∑                      k            =            0                                n                                                              (                                      n              k                                      )                                      =                  2                      n                                                                ∑                      k            =            0                                n                          (        −        1                  )                      k                                                              (                                      n              k                                      )                                      =        0        ,                   where                 n        >        0                                              ∑                      k            =            0                                n                                                              (                                      k              m                                      )                                      =                                            (                                                      n                +                1                                            m                +                1                                                    )                                                                            ∑                      k            =            0                                n                                                              (                                                      m                +                k                −                1                            k                                      )                                      =                                            (                                                      n                +                m                            n                                      )                                               (see Multiset)                              ∑                      k            =            0                                n                                                              (                                      α              k                                      )                                                                          (                                      β                              n                −                k                                                    )                                      =                                            (                                                      α                +                β                            n                                      )                                               (see Vandermonde identity)Sums of sines and cosines arise in Fourier series.
                              ∑                      k            =            1                                ∞                                                              sin                            (              k              θ              )                        k                          =                                            π              −              θ                        2                          ,        0        <        θ        <        2        π                                                              ∑                      k            =            1                                ∞                                                              cos                            (              k              θ              )                        k                          =        −                              1            2                          ln                (        2        −        2        cos                θ        )        ,        θ        ∈                  R                                                                      ∑                      k            =            0                                ∞                                                              sin                            [              (              2              k              +              1              )              θ              ]                                      2              k              +              1                                      =                              π            4                          ,        0        <        θ        <        π                                                              B                      n                          (        x        )        =        −                                            n              !                                                      2                                  n                  −                  1                                                            π                                  n                                                                              ∑                      k            =            1                                ∞                                                1                          k                              n                                                    cos                          (          2          π          k          x          −                                                    π                n                            2                                )                ,        0        <        x        <        1                                                              ∑                      k            =            0                                n                          sin                (        θ        +        k        α        )        =                                            sin                                                                                  (                    n                    +                    1                    )                    α                                    2                                            sin                            (              θ              +                                                                    n                    α                                    2                                            )                                      sin                                                              α                  2                                                                                                                          ∑                      k            =            1                                n            −            1                          sin                                                    π              k                        n                          =        cot                                      π                          2              n                                                                                            ∑                      k            =            1                                n            −            1                          sin                                                    2              π              k                        n                          =        0                                                              ∑                      k            =            0                                n            −            1                                    csc                      2                                            (          θ          +                                                    π                k                            n                                )                =                  n                      2                                    csc                      2                                  (        n        θ        )                                                              ∑                      k            =            1                                n            −            1                                    csc                      2                                                                      π              k                        n                          =                                                            n                                  2                                            −              1                        3                                                                                ∑                      k            =            1                                n            −            1                                    csc                      4                                                                      π              k                        n                          =                                                            n                                  4                                            +              10                              n                                  2                                            −              11                        45                                                                                ∑                      m            =            b            +            1                                ∞                                                b                                          m                                  2                                            −                              b                                  2                                                                    =                              1            2                                    H                      2            b                                  An infinite series of any rational function of                     n                 can be reduced to a finite series of polygamma functions, by use of partial fraction decomposition. This fact can also be applied to finite series of rational functions, allowing the result to be computed in constant time even when the series contains a large number of terms.