# 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.

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.
To add to the other answers pertaining to your specific case, it is important to note that your condition isn't strong enough to define a Cauchy sequence (which is implied by the other answers, but I think should be made explicit). Consider the sequence $(b_n)$ defined by the partial sums of the harmonic series $$b_n=\sum_{k=1}^n\frac{1}{k}.$$ Then $$|b_{n+1}-b_{n}|=\frac{1}{n+1}\longrightarrow 0.$$ However we know that the harmonic series does not converge and therefore $(b_n)$ cannot be Cauchy. If we were given some condition that implied that your $(a_n)$ was, in fact, Cauchy then we would know, by the completeness of $\mathbb{R}$ and uniqueness of limits, that $(a_n)$ converged to $g$ .
Let $$f(x)=(-1)^n \sin (\frac{\pi}{2^n}x), x \in [2^n, 2^{n+1}], n \ge 0.$$ Then $f(x)$ is continuous and vanishes at each point $x=2^n$. Let $a_n = f(n)$, then $a_{2^n}=0, \forall n \ge 0$, and it is easy to verify that $|a_{n+1}-a_n| \rightarrow 0$ as $n \rightarrow \infty$. But $\{a_n\}$ does not converge since $$|a_{k}| = 1, \text{for} \,\,k=\frac{2^{n+1}-2^n}{2},n\ge 1.$$
• One thing I don't quite get is why you consider $|a_{n+1}-a_n|$ as $n \to 0$ instead of $n \to \infty$? – Zelazny Nov 9 '16 at 10:31