Does $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent? Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?
When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$
I have a gut feeling this statement is false, just can't find a series which contradicts it.
Thanks a lot
 A: A counter-example is $a_ n = 2 + \sin(\log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
$$
 \left\lvert \frac{a_{n+1}}{a_n} - 1 \right\rvert 
= \left\lvert \frac{\sin(\log(n+1))- \sin(\log(n))}{2 + \sin(\log(n))} \right\rvert \\
\le \left\lvert \sin(\log(n+1))- \sin(\log(n)) \right\rvert 
= \left\lvert \frac{\cos(\log(x_n))}{x_n}\right\rvert 
$$
for some $x_n \in (n, n+1)$, using the mean value theorem. It follows that
$$
\left\lvert \frac{a_{n+1}}{a_n} - 1 \right\rvert \le \frac{1}{n} \to 0 \, ,
$$
i.e. $\frac{a_{n+1}}{a_n} \to 1$.
Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
so that the ratio of successive sequence elements approaches one.
(This example is taken from $\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = 1 \Rightarrow \exists c \in \mathbb{R}: \lim\limits_{n \to \infty} {a_n} = c$. I copied it here because the other question excludes sequences converging to $\pm \infty$, so it is not an exact duplicate.)
A: A counterexample: $a_n = \exp(\sin\sqrt{n})$. The ratio $\frac{a_{n+1}}{a_n}$ is
$$\frac{a_{n+1}}{a_n}=\exp(\sin(\sqrt{n+1})-\sin(\sqrt{n}))\approx \exp\frac{\cos(\sqrt{n})}{2\sqrt{n}}\approx 1+\frac{\cos(\sqrt{n})}{2\sqrt{n}}\to 1$$
but $a_n$ oscillates between $e^1$ and $e^{-1}$, not converging to anything or diverging to $\infty$.
A: Hint
Think to a sequence $(a_n)$ such that $\lim\limits_{n \to \infty} (a_{n+1}-a_n)=0$, but that oscillates infinitely between $1$ and $2$.
The limit points of $(a_n)$ is the segment $[1,2]$ and $\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}=1$.
Some examples here and there.
