TITLE: Proof that a sequence $(x_n)_{n \ge 1}$ such that $|x_{n+1} - x_n| < \frac{1}{n}$ converges.
QUESTION [6 upvotes]: I'm studying for a test and I came across this exercise:
Let $(x_n)_{n \ge 1}$ be a sequence such that
$x_{2n-1} \le x_{2n+1} \le x_{2n+2} \le x_{2n}, \hspace4ex |x_{n+1} - x_n| < \frac{1}{n}$
Prove that the sequence converges.

I've been stuck on this question for a while now, by doing some arithmetics I've gotten a few inequalities, namely:
(1) $x_{2n} - x_{2n+1} < \frac{1}{2n}$
(2) $x_{2n+2} - x_{2n+1} < \frac{1}{2n+1}$
(3) $x_{2n} - x_{2n-1} < \frac{1}{2n-1}$
But I have no idea where to go from here, or even if those inequalities are useful at all.

REPLY [2 votes]: Argument may need some minimal corrections We have the sequence $x_{2n+1}$ increasing and bounded above by $x_2$, you could try by induction as $x_1\leq x_3\leq x_5\leq x_7\leq x_9 \leq x_{10}\leq x_8\leq x_6\leq x_4\leq x_2$, so for any $n$, $x_{2n+1}\leq x_2$. Then by the Bolzano theorem, it has a subsequence (the odd subsequence has a subsequence $y_{{n}_k}$ which converges to $X$). Then if $x_{2n+1}$ is convengent it should converges to $X$. Notice that $y_{n_k}=x_{2n+1+2p}$ for some $p(n)=p>0$, and $$|x_{2n+1}- X|\leq |x_{2n+1}-y_{n_k}|+|y_{n_k}-X|$$ here you know how to make $|y_{n_k}-X|$ small, it remains to deal with $$|x_{2n+1}-y_{n_k}|=|x_{2n+1}-x_{2n+2}+x_{2n+2}-x_{2n+3}+x_{2n+3}-\cdots+x_{2n+1+2(p-1)}-x_{2n+1+2p}|\leq \frac{2p}{2n+1}\to 0$$ then the next step is to notice $|x_{2n+1}-x_{2n}|<\frac{1}{2n}\to 0$ then $x_{2n}\to X$, so $x_{n}$ also converges to $X$. You could try to do this more rigurous.