TITLE: If $\lim_{n\to\infty} (a_{n+1}-a_n)=0$ and $|a_{n+2}-a_n|<\frac{1}{2^n}$ then $(a_n)$ converges
QUESTION [8 upvotes]: Let $(a_n)$ be a sequence such that $\lim_{n\to\infty} (a_{n+1}-a_n)=0$ and $|a_{n+2}-a_n|<\frac{1}{2^n}$ for all $n$. I have to decide whether or not $(a_n)$ converges.
My attempt: I think it converges. Let $b_n=a_{2n}, c_n=a_{2n-1}$. Then:
$$|b_{n+1}-b_n|=|a_{2n+2}-a_{2n}|<\frac{1}{2^{2n}}$$
$$|c_{n+1}-c_n|=|a_{2n+1}-a_{2n-1}|<\frac{1}{2^{2n-1}}$$
Thus $(b_n)$ and $(c_n)$ are Cauchy (proven in another question) and converge. Because $(a_{2n}-a_{2n-1})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to $0$. Thus $$\lim_{n\to\infty} (b_n-c_n)=\lim_{n\to\infty} (a_{2n}-a_{2n-1})=0$$
or
$$\lim_{n\to\infty} b_n=\lim_{n\to\infty}c_n$$
Because the subsequences $(b_n)$ and $(c_n)$ cover the sequence $(a_n)$ and because they converge to the same point, $(a_n)$ converges.
Is it correct? What do you think?

REPLY [2 votes]: Here you have a sufficient condition (from the discussion in the comments) for your sequence to be convergent: Suppose that $|a_n-a_{n+1}|\leq c_n$, where $\{c_n\}$ is such that
$$
\sum_{n=1}^{\infty}c_n<\infty.
$$
We define the sequence of positive numbers $C_n=\sum_{k=1}^nc_n$, which is increasing. Since it converges, it is also Cauchy. Now we prove that $a_n$ is Cauchy: let $m> n$
$$
|a_n-a_m|=|a_n-a_{n+1}+a_{n+1}-\cdots -a_m|\leq \sum_{k=n}^{m-1} |a_k-a_{k+1}|\leq \sum_{k=n}^{m-1} c_k = C_m-C_{n-1} \to0 
$$
as $m,n\to \infty$.
Sorry for off-topic but the discussion in the comments was quite rich in my opinion, worth to be completed.
EDIT: Actually we only need the summability of the sequence $\{|a_n-a_{n+1}|\}$. Such sequences $\{a_n\}$ are called of bounded variation. So, supposing
$$
\sum_{n=1}^{\infty}|a_n-a_{n+1}|<\infty,
$$
then it suffices to take $c_n=|a_n-a_{n+1}|$ in the previous argument.