Ratio test when checking the convergence of series Suppose we have the series $\sum a_n$.  Define,

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

Then,


if $L&lt;1$ the series is absolutely convergent (and hence convergent).
if $L&gt;1$ 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
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 
$$a_n = \begin{cases}
\frac{1}{2}^n &amp; \text{if $n$ is odd} \\
\frac{1}{3}^n &amp; \text{if $n$ is even} \\
\end{cases}$$
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
$$a_n = \begin{cases}
\frac{1}{2}^n &amp; \text{if $n$ is odd} \\
3^n &amp; \text{if $n$ is even} \\
\end{cases}$$
Clearly if $a_n=0$ for every $n\in\mathbb{N}$, the limit is undefined but the series still converges.