# Convergence of $a_{0} = 0, a_{n}=f(a_{n-1})$ when $|f'(x)|\leq \frac{5}{6}$

By the mean value theorem it's easy to show that $|a_{n+1}-a_{n}| \leq \frac{5}{6}|a_{n}-a_{n-1}|$ for every n.

Next, I thought of saying $|a_{n+1}-a_{n}| \leq ... \leq (\frac{5}{6})^{n}|a_{1}| \to 0$ and somehow show that ** if $M_{n}$ is the closed interval whose end points are $a_{n}$ and $a_{n-1}$ then $a_{n+1} \in M_{n}$ which implies $M_{n+1} \subseteq M_{n}$ and then to finish with Cantor's intersection theorem that gives us convergence of $a_{n}$.

But I'm not even sure if ** is correct and I haven't even used the fact that $a_{0} = 0$.

EDIT: Following the tip and some more thought I've come up with the following:

For every $m\gt n$: $|a_{m}-a_{n}|\leq|a_{m}-a_{m-1}+a_{m-1}-...+a_{n-1}-a_{n}|\leq$
$\leq\sum_{k=n}^{m-1}|a_{k+1}-a_{k}|\leq|a_{1}|\sum_{k=n}^{m-1}(\frac{5}{6})^{k}\le$
$\le|a_{1}|\sum_{k=n}^{\infty}(\frac{5}{6})^{k}=|a_{1}|\frac{(\frac{5}{6})^{n}}{\frac{1}{6}}=6|a_{1}|(\frac{5}{6})^{n} \to 0$ and from here it's easy to show that the sequence is Cauchy.

Are you missing some condition on $f'(x)$? For instance take $f(x) = -100x + 1$. Perhaps $f'(x) \ge 0$? – Aryabhata Nov 12 '10 at 15:27
You presumably want $|f'(x)|\le 5/6$. It's not the case that $a_{n+1}$ need lie in the interval between $a_{n-1}$ and $a_n$. What can you say about $|a_n-a_m|$? (The value of $a_0$ isn't relevant).