an example of a sequence $(u_n)_n$ taking its values in $[-1,+1]$ such that $(u_{n+1}-u_n)$ converge to zero but $(u_n)_n$ does not converge Define a sequence $(u_n)_n$ by:
$$u_n=\cos(\log n).$$
Then, it is easy to show that $(u_{n+1}-u_n)$ goes to zero at infinity.
The question is how to prove that $(u_n)_n$ is a divergent sequence using the subsequence 
$$
    (u_{\varphi(n)})_n,\, \varphi(n)=[e^{\alpha+2\pi n}], \alpha\in[0,\pi].
   $$
(the pair of brackuets $[\cdot]$ denote the integer part function)

Let $n(k)=\lfloor\mathrm e^{2k\pi}\rfloor$ for every positive integer $k$, then $$2k\pi+\log(1-\mathrm e^{-2k\pi})\leqslant\log n(k)\leqslant2k\pi,$$ hence $u_{n(k)}\geqslant\cos(\log(1-\mathrm e^{-2k\pi}))$. Since $\log(1-\mathrm e^{-2k\pi})\to0$, $u_{n(k)}\to+1$ when $k\to\infty$.
Let $m(k)=\lfloor\mathrm e^{(2k+1)\pi}\rfloor$ for every positive integer $k$, then $$2k\pi+\pi+\log(1-\mathrm e^{-(2k+1)\pi})\leqslant\log m(k)\leqslant2k\pi+\pi,$$ hence $u_{m(k)}\leqslant-\cos(\log(1-\mathrm e^{-2k\pi}))$. Since $\log(1-\mathrm e^{-2k\pi})\to0$, $u_{m(k)}\to-1$ when $k\to\infty$.
This shows that $+1$ and $-1$ are limit points of $(u_n)$ (actually the set of limit points of $(u_n)$ is the whole interval $[-1,+1]$), hence the sequence $(u_n)$ diverges.
