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 $\lim\limits_{n\to+\infty}(\frac{1}{u_n^2}-\frac{1}{u_{n-1}^2})$
 A: 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.
$\text{(iv) } |u_n| \leqslant 1/n\text{ for all } n < K\text{.}\\\text{Therefore the limit is 0.}\\\text{I let the OP to complete the details. (to prove (i) and (ii)).}\\\text{Q.E.D. (Chris)}$

A: 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 $u_{n+1}<0$.
