$|s_{2k}-s_k|<\epsilon$ implies $\{s_k\}$ converges? $\{s_k\}$ is a sequence in $\mathbb{R}$.
$\forall\epsilon>0$, $\exists N\in\mathbb{N}$ s.t. $k\geq N$ implies $|s_{2k}-s_k|<\epsilon$.Then, does $\{s_k\}$ converge or not?
I need a proof. Thanks.

It may not converge. For example, let $s_k=1$ when $k=2^n$ for some $n\ge 0$ and let $s_k=0$ otherwise. Then $s_{2k}-s_k=0$ for every $k\ge 1$, and clearly $s_k$ does not converge.
Edit: Even if $s_k$ is nonnegative and increasing, it could also diverge to $+\infty$. For example, let $s_1=0$ and $s_k=\log(\log (k))$ for every $k\ge 2$. Then when $k\ge 2$,
$$s_{2k}-s_k=\int_{\log k}^{\log (2k)}\frac{dx}{x}<\frac{\log 2}{\log k}\to 0, $$
as $k\to\infty$. However, $\lim_{k\to\infty}s_k=+\infty$.
