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Suppose we have the series $\sum a_n$. Define,

$$ L=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} $$

Then,

  • if $L<1$ the series is absolutely convergent (and hence convergent).
  • if $L>1$ the series is divergent.
  • if $L=1$ the series may be divergent, conditionally convergent, or absolutely convergent.

What if $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}$ doesn't exist, does it mean that series $\sum a_n$ diverges? I think that if sometimes our ratio test doesn't work, then by checking other tests we might find that series converges. Am I right?

If I am not right, can you show me the proof. Thanks

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You can't conclude anything from the fact $\lim_{n \to \infty} a_{n+1}/a_n$ does not exist.

The series may converge in that case: for example, let $$a_n = \begin{cases} \frac{1}{2}^n & \text{if $n$ is odd} \\ \frac{1}{3}^n & \text{if $n$ is even} \\ \end{cases}$$ In this case, $\liminf a_{n+1}/a_n = 2/3$ and $\limsup a_{n+1}/a_n = 3/2$, so $\lim a_{n+1}/a_n$ does not exist. But the series converges because $0 \leq a_n \leq 1/2^n$ for all $n$.

On the other hand, it's also possible for the series to diverge: for example, let $$a_n = \begin{cases} \frac{1}{2}^n & \text{if $n$ is odd} \\ 3^n & \text{if $n$ is even} \\ \end{cases}$$

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Clearly if $a_n=0$ for every $n\in\mathbb{N}$, the limit is undefined but the series still converges.

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  • $\begingroup$ What was wrong with this? Surely all you need is 1 counterexample... $\endgroup$ – Sam Spedding May 15 '18 at 18:38

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