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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 $0<x_1<1, x_{n+1}=\sin x_{n}$, 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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The expression $$\lim_{n\to\infty}x_n=\frac{2}{n}$$makes no sense: $\,n\,$ cannot appear on the RHS. –  DonAntonio Jul 18 '12 at 16:56
    
I've modified the original statement, thanks. –  Benson Jul 18 '12 at 17:29
1  
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? –  Did Aug 16 '12 at 14:52

1 Answer 1

up vote 1 down vote accepted

Since $$f(x)=\log(1+x)-x\,,\,x>0$$ is a decreasing function, we get that $\,f(x)\geq f(y)=f(0)\,\,,\,\forall 0<x<y\,$ , 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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Thank you, @DonAntonio. According to your answer, we can easily get $\lim_{n\to\infty}x_n=0$, however, this seems a bit trival since here I want to prove is $\lim_{n\to\infty}nx_n=2$, which seems harder(The original statement is really misleading, I've modified it) –  Benson Jul 18 '12 at 17:27
1  
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. –  DonAntonio Jul 18 '12 at 20:16

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