Ratio test counterexample

Given a sequence a1, a2, a3, … let L be the limit of the ratio of consecutive terms:

L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n}\right|

Then the series

\sum_{n=1}^\infty a_n

converges if L < 1 and diverges if L > 1.

However, that’s not the full story. Here is an example from Ernesto Cesàro (1859–1906) that shows the ratio test to be more subtle than it may seem at first. Let 1 < α < β and consider the series

\frac{1}{1^\alpha} + \frac{1}{2^\beta} + \frac{1}{3^\alpha} + \frac{1}{4^\beta} + \cdots

The ratio a2n + 1 / a2n diverges, but the sum converges.

Our statement of the ratio test above is incomplete. It should say if the limit exists and equals L, then the series converges if L < 1 and diverges if L > 1. The test is inconclusive if the limit doesn’t exist, as in Cesàro’s example. It’s also inconclusive if the limit exists but equals 1.

Cesàro’s example interweaves two convergent series, one consisting of the even terms and one consisting of the odd terms. Both converge, but the series of even terms converges faster because β > α.

Related post: Cesàro summation

4 thoughts on “Ratio test counterexample

  1. This doesn’t look especially subtle to me. The ratio test asks for the limit of a_{n+1} / a_n . Then the “counterexample” wants to talk about the limit of a completely different ratio. And the limit of that ratio turns out to be irrelevant to the ratio test. And…?

    The original statement (“let L = lim_{n \to \infty} | a_{n+1} / a_n |; the series sum(a_n) converges if L < 1 and diverges if L > 1”) isn’t incomplete at all; when L does not exist, we cannot claim that L < 1 or that L > 1.

  2. Juan José Alba González

    I agree with Michael Watts, the existence of the limit is implicit since one set $L = \lim_{n\to\infty} $…

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