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

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    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.2012-09-16

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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.
(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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    @uforoboa: maybe you wanna read this version (posted by my brother).2012-09-15
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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. 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 for some $n$.

Let $q_n = {u_n\over n}$ for $n\ge 1$. We have seen that $0 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.2012-09-03