# Find $\lim\limits_{n\to+\infty}(u_n\sqrt{n})$

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 $\lim\limits_{n\to+\infty}(\frac{1}{u_n^2}-\frac{1}{u_{n-1}^2})$

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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 3 '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

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.
(iv) |u_n| <= 1/n for all n < K."

Therefore the limit is $0$.

I let the OP to complete the details. (to prove (i) and (ii)).

Q.E.D. (Chris)

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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 3 '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 3 '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 1618033 Sep 3 '12 at 17:37
@Chris.. this is an essential point missing in your answer.. if it is evident for you then add it to the answer.. – uforoboa Sep 3 '12 at 18:58
Moreover to me it is not evident how to use Cesaro-Stolz in your argument.. – uforoboa Sep 3 '12 at 19:00

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 $u_{n+1}<0$.

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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 3 '12 at 6:59