Is this implication true? Suppose that a real sequence $u_n$ is such that $$u_{n+1}-u_n \rightarrow0$$
That is not enough to prove that $u_n$ is convergent (take $u_n=ln(n)$)
Now what if $u_n$ is bounded ? I guess it does converge, but how to prove this ? I tried to show that it had only one accumulation point... 
 A: Bounded does not help. Use the sequence made up of partial sums from the sequence
$$1, -\frac{1}{2},-\frac{1}{2}, \frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}, -\frac{1}{8}, -\frac{1}{8},-\frac{1}{8},-\frac{1}{8},-\frac{1}{8},-\frac{1}{8},-\frac{1}{8},-\frac{1}{8},\dots.$$
A: The sequence
$$0, 1, \frac{1}{2}, 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1, \frac{7}{8},\frac{6}{8},\frac{5}{8}, ..., \frac{1}{8}, 0 , \frac{1}{16}, \frac{2}{16}, ..., 1 , \frac{31}{32}, ...$$
is not convergent and it has infinitely many accumulation points (every $x\in [0,1]$ is accumulation point).
A: Hint: Consider
$$
u_n=\sin(\log(n))
$$
Show that $|u_{n+1}-u_n|\lt\frac1n$ yet $\limsup\limits_{n\to\infty}u_n=1$ and $\liminf\limits_{n\to\infty}u_n=-1$.
A: In $\mathbf R$ it won't be a sufficient condition. But if you want to see spaces where this condition is really useful, you can look about ultrametric spaces. It's the name for metric spaces $(X,d)$ where $\forall (x,y,z)\in X^3, d(x,y)\leq\sup(d(x,z),d(z,y))$ (an stronger inequality of triangle). In this spaces a sequence is a Cauchy sequence if and only if $d(u_n,u_{n+1})\rightarrow 0$. So in a complete ultrametric space, a sequence is convergent if and only if $d(u_n,u_{n+1})\rightarrow 0$.
I notice now, that in $\mathbf R$, the sequence $u_n$ is convergent if and only if the serie $\sum (u_n- u_{n-1})$ is convergent (for example if $u_n-u_{n-1}=O(1/n^a)$ where $a>1$.
