Sequence convergence Cauchy and subsequence

Question

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 1

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.

Answer 2

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 $g$ .

Answer 3

I have constructed another counter-example:

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 $$|a_{k}| = 1, \text{for} \,\,k=\frac{2^{n+1}-2^n}{2},n\ge 1.$$