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Let ${u_n}$ be a sequence defined by $u_o=a \in [0,2), u_n=\frac{u_{n-1}^2-1}{n} $ for all $n \in \mathbb N^*$ Find $\lim\limits_{n\to+\infty}{(u_n\sqrt{n})}$

I try with Cesaro, find $\lim\limits_{n\to+\infty}(\frac{1}{u_n^2}-\frac{1}{u_{n-1}^2})$ then we get $\lim\limits_{n\to+\infty}(u_n^2n)$ But I can't find $\lim\limits_{n\to+\infty}(\frac{1}{u_n^2}-\frac{1}{u_{n-1}^2})$

tangkhaihanh
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  • Since you already accepted an answer, you might want to complete it, answering the points raised in the comments. – Did Sep 03 '12 at 17:33
  • The very same comment applies to the entirely revised version of the accepted answer. Note that (iv) and the first part of (iii) are not obviously related to the rest, that (iv) is false, and that nothing in (i)-(iv) implies that the limit of $u_n\sqrt{n}$ is zero. – Did Sep 16 '12 at 12:32

2 Answers2

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A solution from a friend of mine:

(i) Show $u_{n} > -1$ for all $n$. (Easy)
(ii) If $u_{0} = 2 - 2t$, where $0 \le t \le 1$ then $u_{n} < (n+2)(1-t)$ for all $n > 0$. (Induction)
(iii) There exists integer $K > 0$ s.t. $-1 < u_K < 1$. From (ii) we get that eventually $u_{n-1} < n$, whence $u_n < n$, and $u_{n+1} < n-1 $, etc.
$\text{(iv) } |u_n| \leqslant 1/n\text{ for all } n < K\text{.}\\\text{Therefore the limit is 0.}\\\text{I let the OP to complete the details. (to prove (i) and (ii)).}\\\text{Q.E.D. (Chris)}$

Angelo
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user 1591719
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    Two questions.. it seems that you are assuming that $\lim u_n$ exists finite. I don't find it so obvious so can you add details? It depends heavily on $a\in [0, 2)$.the other is.. how is the last equality in your long chain valid? – uforoboa Sep 03 '12 at 13:37
  • To add to, and partially repeat, @uforoboa's point, note that if $u_0=2$, then $u_n=n+2$ for every $n$. Hence $u_n\sqrt{n}$ diverges for every $a\geqslant2$. Since the proof above uses nowhere the initial value $u_0$, something is lacking. Furthermore, it seems that the first application of Stolz-Cesàro is going in the wrong direction. – Did Sep 03 '12 at 17:30
  • @ did: "it seems" or it is wrong? Which one? If $u_o=a \in [0,2)$ then $u_n$ converges. – user 1591719 Sep 03 '12 at 17:37
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    @Chris.. this is an essential point missing in your answer.. if it is evident for you then add it to the answer.. – uforoboa Sep 03 '12 at 18:58
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    Moreover to me it is not evident how to use Cesaro-Stolz in your argument.. – uforoboa Sep 03 '12 at 19:00
  • @uforoboa: please let me know where I'm wrong. I just want to learn. I admit that I may be wrong. Maybe I miss something. Thanks. – user 1591719 Sep 03 '12 at 19:20
  • @uforoboa: I'll try to improve that point you talked about at first. – user 1591719 Sep 03 '12 at 19:26
  • Chris'sister: Since you ask, "it seems" is a polite version of "this is obviously wrong". The second sentence of your comment to me is a joke, just read the remark I made. (Unrelated: Using "@ did" with a space between "@" and "did" is pretty efficient to make sure your comment is NOT notified to me.) – Did Sep 09 '12 at 15:16
  • @did: let's see what is wrong, which part and I'll be glad to correct it. Everybody makes mistakes and I do, but let's firstly see the mistake. – user 1591719 Sep 09 '12 at 16:33
  • @Chris'ssister Begin of quote: Since the proof above uses nowhere the initial value $u_0$, something is lacking. End of quote. Begin of quote: The first application of Stolz-Cesàro is going in the wrong direction. End of quote. (Are you reading the comments made to you?) – Did Sep 09 '12 at 16:40
  • @did: what do you mean by "The first application of Stolz-Cesàro is going in the wrong direction"?. What is wrong? and why is wrong? In what sense? – user 1591719 Sep 09 '12 at 16:45
  • @Chris'ssister Do you even know what Stolz-Cesàro says? Any comment about the relevance of $u_0$? // Anyway, after your deliberately insulting comments on the other page (for which you should get suspended), I suddenly feel the urge to do something else with my time than to try to help you. – Did Sep 09 '12 at 16:51
  • I'll try to improve this answer as I previously said. – user 1591719 Sep 09 '12 at 17:06
  • @uforoboa: maybe you wanna read this version (posted by my brother). – user 1591719 Sep 15 '12 at 19:37
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If ever $u_N\le 0$, then all $-1/n\le u_n\le0$ for all $n>N$, hence $u_n\sqrt n\to 0$. Therefore we may assume for the rest of the argument that $u_n>0$ for all $n$.

Let $e_n = n+2-u_n$. Then $0<e_0<2$. Using the recursion formula for $e_n$ show that the assumption that $e_n\le2$ for all $n$ leads to $e_n\ge2^n e_0$. Therefore $e_n>2$ for some $n$, i.e. $u_n<n$ for some $n$.

Let $q_n = {u_n\over n}$ for $n\ge 1$. We have seen that $0<q_n<1$ for big $n$. Find the recursion formula for $q_n$ and show that $q_n< q_{n-1}^2$ for big $n$ and therefore $q_n<\frac1n$ for some $n$. But then $u_{n+1}<0$.

Hagen von Eitzen
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  • Two points: (1.) Some details might be lacking in the proof that $e_n\geqslant2^ne_0$ for every $n$ if $e_n\leqslant2$ for every $n$. (2.) It seems that one shows that $e_n\gt2$ for at least one $n$ but then one uses that $e_n\gt2$ for every $n$ large enough. – Did Sep 03 '12 at 06:59