TITLE: Prove that a recurrent sequence $u_n$ is convergent if $\lim \limits_{n \to \infty}(u_{n+1}-u_n)=0$
QUESTION [9 upvotes]: Let $f:[a;b] \to [a;b]$ be a continuous function ($a$ and $b$ real numbers) and $(u_n)$ a sequence defined by  $u_0 \in [a;b]$ and $u_{n+1}=f(u_n)$
Prove that if $\lim \limits_{n \to \infty}\left(u_{n+1}-u_n\right)=0$ then $(u_n)$ converges
I tried using subsequences and the bolzano weierstrass theorem but couldn't even get close.
Any help would be welcome.

REPLY [3 votes]: $\{u_n\}$ is a sequence taking values in the compact segment $[a,b]$. Hence it has a convergent subsequence $\{u_{\varphi(n)}\}$ where $\varphi : \mathbb N \to \mathbb N$ is strictly increasing. Let say that $\lim\limits_{n \to \infty} u_{\varphi(n)} = l \in [a, b]$.
Suppose that $\{u_n\}$ has a second limit point $L \neq l$. We can find a subsequence $\{u_{\psi(n)}\}$ converging to $L$. Without loss of generality, we can suppose $l \lt L$. As $\lim \limits_{n \to \infty}\left(u_{n+1}-u_n\right)=0$, any $ x \in [l , L]$ is a limit point of $\{u_n\}$.
Now as $f$ is supposed to be continuous, we get that $f(x) = x$ for all $x \in [l,L]$. But if $u_N \in (l,L)$ for $N \in \mathbb N$, we get the contradiction $u_p = u_N$ for all $p \ge N$ and therefore that $u_N$ is the only limit point of $\{u_n\}$.
Conclusion: $\{u_n\}$ has a single limit point and therefore converges.