<s>[INST] 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 : $$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$$
 [/INST] 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
$$\lim_{n\to\infty} \left ( \frac{u_{n+1}}{{u_{1}}{u_{2}}\cdots{u_{n}}} \right )^2 = 2011.$$
</s>