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How to prove the convergence for function of function series?

Say, here're two examples

  • Given $x_1>0, x_{n+1}=\ln(1+x_n)$, Prove $\lim_{n\to\infty}nx_n=2$
  • Given $0, Prove $\lim_{n\to\infty}\sqrt{n}x_n$ exist, and give this limit.

I've written programs to check the above two problems, and it seems the assertions are true, however I found proving this kind of problems extreamly hard, since simply expanding this function of function series usually make me totally lost. Any one can give me some suggestions on such problems? Thanks a lot!

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    Benson: I am not sure that you understand what is going on on this page, so let me explain. You asked a question, to which @DonAntonio fully answered. You then modified the question, making Don's answer appear irrelevant. These are not correct practices and I, for one, would not want to answer the revised version of your question, not wishing to condone them. Your option here is to go back to the original version of your post, (probably) to accept Don's answer, and to post your new question as a new post... which will then receive tons of illuminating answers. Nice, eh?2012-08-16

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Since $f(x)=\log(1+x)-x\,,\,x>0$ is a decreasing function, we get that $\,f(x)\geq f(y)=f(0)\,\,,\,\forall 0 , and it follows from here that the sequence $\,\{x_n\}=\{\log(1+x_{n-1})\}\,$ is decreasing and obviously bounded from below by say $\,\log 1 =0\,$ , so the sequence's limit exist and you can find with arithmetic of limits and using the continuity of $\,\log x\,$: $\text{if}\,\,\alpha=\lim_{n\to\infty}x_n\,\,,\,\,\alpha=\lim_{n\to\infty} x_{n+1}=\lim_{n\to\infty}\log(1+x_n)=\log(1+\alpha)\Longrightarrow$

$\Longrightarrow e^\alpha=1+\alpha\Longrightarrow e^\alpha-\alpha-1=0$

Do you recognize $\,\alpha\,$? :)

Try the second one on the same lines as above (decreasing sequence and etc.)

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    Well, after you modified the statement of your question my answer helps only yo find the limit of $\,\{x_n\}\,$ ...I cannot guess when somebody is going to change his own question *after* getting answers to his OP.2012-07-18