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 : $$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$$
 A: 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 $n \rightarrow \infty$:
$$=\frac{(2015-2)^2}{2015}-\frac{4}{2015}+0$$
$$=\frac{2015^2-4\cdot 2015 +4}{2015}-\frac{4}{2015}$$
$$=2015-4=2011$$
A: \begin{aligned}
u_{n+1} = {u_{n}}^2 - 2 \\
{u_{n+1}}^2 = ({u_{n}}^2 - 2)^2 \\
{u_{n+1}}^2 = {u_{n}}^4 - 4{u_{n}}^2 + 4 \\
{u_{n+1}}^2 - 4 = {u_{n}}^2({u_{n}}^2 - 4) \\
{u_{n}}^2 = \frac{{u_{n+1}}^2 - 4}{{u_{n}}^2 - 4}
\end{aligned}
A: The recurrence relation may be expressed as
$$ 
{u_{n}}^2 = \frac{{u_{n+1}}^2 - 4}{{u_{n}}^2 - 4}.
$$
Which allows telescopic cancellation,
$$
\begin{align*}
\left ({u_{1}}{u_{2}}\cdots{u_{n}}\right )^2 &=
\left (\frac{\color{#036}{{u_{2}}^{2} - 4}}{{u_{1}}^2 - 4} \right )
\left ( \frac{\color{#063}{{u_{3}}^{2} - 4}}{\color{#036}{{u_{2}}^2 - 4}} \right )
\cdots
\left ( \frac{{u_{n+1}}^{2} - 4}{\color{#630}{{u_{n}}^2 - 4}} \right )\\[10pt]
&= \frac{{u_{n+1}}^2 - 4}{{u_{1}}^2 - 4}.
\end{align*}
$$
Hence,
$$
\begin{align*}
\lim_{n\to\infty} \left ( \frac{u_{n+1}}{{u_{1}}{u_{2}}\cdots{u_{n}}} \right )^2
&= \lim_{n\to\infty} {u_{n+1}}^2 \cdot \frac{{u_{1}}^2 - 4}{{u_{n+1}}^2 - 4}\\
&= \lim_{n\to\infty} \left ( {u_{1}}^2 - 4\right ) \cdot \frac{1}{1 - 4/{u_{n+1}}^2}\\
&= {u_{1}}^2 - 4 &\text{since $\lim_{n\to\infty} 1/u_{n+1} = 0$}.
\end{align*}
$$
Substituting $u_1 = \sqrt{2015}$ yields the final result
$$\lim_{n\to\infty} \left ( \frac{u_{n+1}}{{u_{1}}{u_{2}}\cdots{u_{n}}} \right )^2 = 2011.$$
