If $a_{1}>2$and $a_{n+1}=a_{n}^{2}-2$ then Find $\sum_{n=1}^{\infty}$ $\frac{1}{a_{1}a_{2}......a_{n}}$ Question If $a_{1}>2$and $\left\{ a_{n}\right\} be$ a recurrsive
sequence defined by setting $a_{n+1}=a_{n}^{2}-2$ then Find
$\sum_{n=1}^{\infty}$$\frac{1}{a_{1}a_{2}......a_{n}}$

Book's Answer
I have mentioned my problem in the image.Any and all help will be appreciated

You have proven that 
$$a_n^2-4 = a_1^2 \ldots a_{n-1}^2(a_1^2-4)$$
Divide both sides by $a_1^2 \ldots a_{n-1}^2$ and then take square root on both sides,
We have 
$$\sqrt{\frac{a_n^2-4}{a_1^2 \ldots a_{n-1}^2}} = \sqrt{a_1^2-4}$$
Now take limit $n \to \infty$, 
\begin{align}\lim_{n \to \infty}\sqrt{\frac{a_n^2-4}{a_1^2 \ldots a_{n-1}^2}}   = \sqrt{a_1^2-4 }& \text{, note that RHS is independent of $n$.}\\
\sqrt{\lim_{n \to \infty}\frac{a_n^2-4}{a_1^2 \ldots a_{n-1}^2}}   = \sqrt{a_1^2-4} & \text{ since square root is continuous}\\
\sqrt{\lim_{n \to \infty}\frac{a_n^2}{a_1^2 \ldots a_{n-1}^2}- \lim_{n \to \infty}\frac{4}{a_1^2 \ldots a_{n-1}^2}}   = \sqrt{a_1^2-4}\\
\sqrt{\lim_{n \to \infty}\frac{a_n^2}{a_1^2 \ldots a_{n-1}^2}}   = \sqrt{a_1^2-4} & \text{, the $a_n$ increases to $\infty$, the second term vanishes}\\
\lim_{n \to \infty}\frac{a_n}{a_1 \ldots a_{n-1}}   = \sqrt{a_1^2-4}
\end{align}
