0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Do you mean limit is $\infty$ $\endgroup$ – Abhishek May 15 '18 at 16:59
  • $\begingroup$ Ratio test is not a nice test $\endgroup$ – Abhishek May 15 '18 at 16:59
  • $\begingroup$ it doesn't work even on $\sum_{n=1}^n\frac{1}{n^2}$ $\endgroup$ – Abhishek May 15 '18 at 17:02
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Abhishek May 15 '18 at 17:05
3
$\begingroup$

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 $$a_n = \begin{cases} \frac{1}{2}^n & \text{if $n$ is odd} \\ 3^n & \text{if $n$ is even} \\ \end{cases}$$

$\endgroup$
-2
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.