When does $\lim\limits_{n\to\infty}|a_{n+1}-a_n|=0$ imply $a_n$ convergent? 
Is it true: Let $\{a_n\}$ be real sequence. If $|a_{n+1}-a_n|<\frac1{3^n}$ for all $n$, then $\{a_n\}$ convergent.

I am asking this, because I was pointed out that there is a real sequence $\{a_n\}$ such that $|a_{n+1}-a_n|<\frac1{n}$, yet $\{a_n\}$ not convergent.

Note that the sequence $(a_n)_n$ converges if, and only if, the series $\sum_n (a_{n+1}-a_n)$ converges. (Can you see why?)
But if we have $$\forall n\,\qquad  \lvert a_{n+1}-a_n \rvert < \frac{1}{3^n}$$
then by comparison with $\sum_n \frac{1}{3^n}$ the series $\sum_n  \lvert a_{n+1}-a_n \rvert$ converges, i.e., the series $\sum_n (a_{n+1}-a_n)$ converges absolutely. And therefore converges.

Now, the thing in this argument which fails when we are only told $\lvert a_{n+1}-a_n \rvert < \frac{1}{n}$ is when comparing to the series $\sum_n \frac{1}{n}$... as this series does not converge.
