limits - Sequence convergence Cauchy and subsequence

I have a sequence $(a_n)$. Its subsequence $a_{2^n}$ converges to a finite limit $g$. The sequence $|a_{n+1} - a_n|$ converges to $0$. Can we conclude about the convergence of $(a_n)$? I suspect I have to notice this is a Cauchy sequence, however, I'm not sure how to use the fact that $a_{2^n}$ converges to $g$.

The Cauchy condition for convergence states that $(a_n)$ converges if and only if
$$\forall \epsilon>0 , \exists n_{\epsilon} \in \mathbb{N} , s.t.\forall m,k > n_{\epsilon}, |a_m - a_k|<\epsilon.$$
Alternatively, we could show that $(a_n)$ is bounded from above and monotonic, however, I can't quite arrive at this conclusion either.

Answer

Consider the sequence given by
$$a_{2^n+k} = \cases{g+\frac{k}{2^{n-1}} & for $0\le k\le2^{n-1}$,\\g+\frac{2^n-k}{2^{n-1}}& for $2^{n-1}\le k\le 2^n$.}$$
Obviously $a_{2^n} = g$, so this subsequence converges, and $|a_m - a_{m+1}| \le 2^{1-n}$, where $n$ biggest natural number such that $2^n\le m$, so it converges to zero. However, $a_{2^n+2^{n-1}} = g+1$ for all $n$, so the sequence is not convergent.

The problem with this sequence is that yes, you have a subsequence giving you some control on the terms, but the element composing the subsequence are spread further and further apart, so that we have enough freedom to make the whole sequence oscillate enough to make it non-convergent.