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

Please correct me if I made an error.

  • $\begingroup$ Are you missing some condition on $f'(x)$? For instance take $f(x) = -100x + 1$. Perhaps $f'(x) \ge 0$? $\endgroup$ – Aryabhata Nov 12 '10 at 15:27
  • $\begingroup$ Or |f'(x)| \le 5/6. In any case, all you need to do is show that a_n is Cauchy. $\endgroup$ – Qiaochu Yuan Nov 12 '10 at 15:40

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).

  • $\begingroup$ Thanks, I believe it helped. See my edit. $\endgroup$ – daniel.jackson Nov 12 '10 at 16:35

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