TITLE: Is this implication true?
QUESTION [7 upvotes]: 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...

REPLY [3 votes]: 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 $a>1$.