Ratio test

Motivation

Given the following geometric series:

The quotient

of any two adjacent terms is 1 2 . The sum of the first m terms is given by:

As m increases, this converges to 1, so the sum of the series is 1. On the other hand given this geometric series:

The quotient a n + 1 a n of any two adjacent terms is 2. The sum of the first m terms is given by

which increases without bound as m increases, so this series diverges. More generally, the sum of the first m terms of the geometric series is given by:

Whether this converges or diverges as m increases depends on whether r, the quotient of any two adjacent terms, is less than or greater than 1. Now consider the series:

This is similar to the first convergent sequence above, except that now the ratio of two terms is not fixed at exactly 1/2:

However, as n increases, the ratio still tends in the limit towards the same constant 1/2. The ratio test generalizes the simple test for geometric series to more complex series like this one where the quotient of two terms is not fixed, but in the limit tends towards a fixed value. The rules are similar: if the quotient approaches a value less than one, the series converges, whereas if it approaches a value greater than one, the series diverges.

The test

The usual form of the test makes use of the limit

The ratio test states that:

It is possible to make the ratio test applicable to certain cases where the limit L fails to exist, if limit superior and limit inferior are used. The test criteria can also be refined so that the test is sometimes conclusive even when L = 1. More specifically, let

Then the ratio test states that:

If the limit L in (1) exists, we must have L = R = r. So the original ratio test is a weaker version of the refined one.

Convergent because L < 1

Consider the series or sequence of series

Putting this into the ratio test:

As every term is positive, the series converges.

Divergent because L > 1

Consider the series

Putting this into the ratio test:

Thus the series diverges.

Inconclusive because L = 1

Consider the three series

The first series (1 + 1 + 1 + 1 + ⋯) diverges, the second one (the one central to the Basel problem) converges absolutely and the third one (the alternating harmonic series) converges conditionally. However, the term-by-term magnitude ratios | a n + 1 a n | of the three series are respectively 1 ,    n 2 ( n + 1 ) 2    and    n n + 1 . So, in all three cases, we have lim n → ∞ | a n + 1 a n | = 1 . This illustrates that when L=1, the series may converge or diverge and hence the original ratio test is inconclusive. For the first series ∑ n = 1 ∞ 1 , however, as the term-by-term magnitude ratio | a n + 1 a n | = 1 for all n, we can apply the third criterion in the refined version of the ratio test to conclude that the series diverges.

Proof

Below is a proof of the validity of the original ratio test.

Suppose that L = lim n → ∞ | a n + 1 a n | < 1 . We can then show that the series converges absolutely by showing that its terms will eventually become less than those of a certain convergent geometric series. To do this, let r = L + 1 2 . Then r is strictly between L and 1, and | a n + 1 | < r | a n | for sufficiently large n (say, n greater than N). Hence | a n + i | < r i | a n | for each n > N and i > 0, and so

That is, the series converges absolutely.

On the other hand, if L > 1, then | a n + 1 | > | a n | for sufficiently large n, so that the limit of the summands is non-zero. Hence the series diverges.

Extensions for L = 1

As seen in the previous example, the ratio test may be inconclusive when the limit of the ratio is 1. Extensions to ratio test, however, sometimes allows one to deal with this case. For instance, the aforementioned refined version of the test handles the case

Below are some other extensions.

Raabe's test

This extension is due to Joseph Ludwig Raabe. It states that if

then the series will be absolutely convergent if R>1 and divergent if R<1. d'Alembert's ratio test and Raabe's test are the first and second theorems in a hierarchy of such theorems due to Augustus De Morgan.

Higher order tests

The next cases in de Morgan's hierarchy are Bertrand's and Gauss's test. Each test involves slightly different higher order asymptotics. Bertrand's test asserts that if

then the series converges if lim inf ρ_n > 1, and diverges if lim sup ρ_n < 1.

Gauss's test asserts that if

where r > 1 and C_n is bounded, then the series converges if h > 1 and diverges if h ≤ 1.

These are both special cases of Kummer's test for the convergence of the series Σa_n, for positive a_n. Let ζ_n be an auxiliary sequence of positive constants. Let

Then if ρ > 0, the series converges. If ρ < 0 and Σ1/ζ_n diverges, then the series diverges. Otherwise the test is inconclusive.

Proof of Kummer's test

If ρ > 0 then fix a positive number 0 < δ < ρ . There exists a natural number N such that for every n > N ,

Since a n + 1 > 0 , for every n > N ,

In particular ζ n + 1 a n + 1 ≤ ζ n a n for all n ≥ N which means that starting from the index N the sequence ζ n a n > 0 is monotonically decreasing and positive which in particular implies that it is bounded below by 0. Therefore the limit

This implies that the positive telescoping series

and since for all n > N ,

by the direct comparison test for positive series, the series ∑ n = 1 ∞ δ a n + 1 is convergent.

On the other hand, if ρ < 0 , then there is an N such that ζ n a n is increasing for n > N . In particular, there exists an ϵ > 0 for which ζ n a n > ϵ for all n > N , and so ∑ n a n = ∑ n a n ζ n ζ n diverges by comparison with ∑ n ϵ ζ n .