How to prove :$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$ For sequence $u_n$ satisfing : $$\begin{cases} u_1=\sqrt{2015}\\ u_{n+1}=u_n^2-2\end{cases}$$
How to prove : $$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$$

Here I use its properties as an additive telescoping series:
$$\left(\frac{u_n^2-2}{u_1\cdots u_n}\right)^2=\frac{u_n^4}{(u_1\cdots u_n)^2}-\frac{4u_n^2}{(u_1\cdots u_n)^2}+\frac{4}{(u_1\cdots u_n)^2}$$
$$=\frac{u_n^2}{(u_1\cdots u_{n-1})^2}-\frac{4}{(u_1\cdots u_{n-1})^2}+\frac{4}{(u_1\cdots u_n)^2}$$
Substitute again:
$$=\left(\frac{u_{n-1}^2}{(u_1\cdots u_{n-2})^2}-\frac{4}{(u_1\cdots u_{n-2})^2}+\frac{4}{(u_1\cdots u_{n-1})^2} \right)-\frac{4}{(u_1\cdots u_{n-1})^2}+\frac{4}{(u_1\cdots u_n)^2}$$
$$=\frac{u_{n-1}^2}{(u_1\cdots u_{n-2})^2}-\frac{4}{(u_1\cdots u_{n-2})^2}+\frac{4}{(u_1\cdots u_n)^2}$$
And again:
$$=\left(\frac{u_{n-2}^2}{(u_1\cdots u_{n-3})^2}-\frac{4}{(u_1\cdots u_{n-3})^2}+\frac{4}{(u_1\cdots u_{n-2})^2} \right)-\frac{4}{(u_1\cdots u_{n-2})^2}+\frac{4}{(u_1\cdots u_n)^2}$$
$$=\frac{u_{n-2}^2}{(u_1\cdots u_{n-3})^2}-\frac{4}{(u_1\cdots u_{n-3})^2}+\frac{4}{(u_1\cdots u_n)^2}$$
Reapeat this process ad infinitum (noting that $\lim_{n \rightarrow \infty}u_n= \infty$):
$$=\frac{u_{2}^2}{(u_1)^2}-\frac{4}{(u_1)^2}+\frac{4}{(u_1\cdots u_n)^2}$$
And take the limit $n \rightarrow \infty$:
$$=\frac{(2015-2)^2}{2015}-\frac{4}{2015}+0$$
$$=\frac{2015^2-4\cdot 2015 +4}{2015}-\frac{4}{2015}$$
$$=2015-4=2011$$
