6
$\begingroup$

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.

Please correct me if I made an error.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.