Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a sequence $\{x_n\}_n$ and real numbers $c > 0$ and $L$, such that $\displaystyle\lim_{n \to \infty}x_n - L = 0$ and $\displaystyle\lim_{n \to \infty} \frac{| x_{n+1} - L |}{|x_n - L|^p} = c$, prove that $p \geq 1$.

This is assumed without proof in my textbook and I'd like a rigorous one, but I can't come up with it.

share|cite|improve this question
Your question title leads to confusion : in the title you should put $\ge$ instead of $>$. – Patrick Da Silva Feb 6 '12 at 23:39
thanks, updated – asmodius Feb 7 '12 at 0:11
Which textbook is this? Have you provided us the whole context? – Aryabhata Feb 7 '12 at 0:36
You should've mentioned $c > 0$ if this is behind the definition of order of convergence. – Patrick Da Silva Feb 7 '12 at 0:53
thanks again, "Calcolo Numerico" by Brugnano et al. didn't mention this. – asmodius Feb 7 '12 at 1:08
up vote 2 down vote accepted

If we allow $c = 0$, then this is not true.

Take $$x_n = \frac{1}{n}$$

where we can pick $p = 0.5$

If $c \gt 0$, then I believe it is true.

We use the following fact:

If $f_n \to \infty$ and if $\dfrac{f_{n+1}}{f_n} \to p$, then $p \ge 1$.

(because otherwise $\sum f_n$ would be absolutely convergent, by the ratio test!)

Assuming $x_n \neq L$ for any $n$ (otherwise problem is meaningless I suppose).

Picking $f_n = -\log |x_n - L|$ will give the result, as I believe we can then show that $\dfrac{f_{n+1}}{f_n} \to p$.

share|cite|improve this answer
In the context of numerical analysis, we usually assume $c > 0$. Can you still find something? I thought of $1/n$ too but I knew somehow OP just forgot to mention that, but I should've commented it. The idea is that in the hypothesis of OP's question, we have the definition of order of convergence. – Patrick Da Silva Feb 7 '12 at 0:52
@PatrickDaSilva: I see. Thanks. – Aryabhata Feb 7 '12 at 0:59
@PatrickDaSilva: If $c \gt 0$, then it is true I think. I have added a sketch of a proof/approach. – Aryabhata Feb 7 '12 at 1:49
Yes, this is essentially the approach. I went down through my numerical analysis notes and that's similar to the proof I found there. Good work! – Patrick Da Silva Feb 7 '12 at 7:55
@PatrickDaSilva: Thanks! – Aryabhata Feb 7 '12 at 19:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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