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
 A: Note that
$$a_{n}=a_{n-1}^{2}-2\implies a_n^2=a_{n-1}^{4}-4a_{n-1}-4\implies a_n^2-4=a_{n-1}^{2}(a_{n-1}^{2}-4) $$
thus
$$a_{n-1}^2-4=a_{n-2}^{2}(a_{n-2}^{2}-4) \implies a_n^2-4=a_{n-1}^{2}a_{n-2}^{2}(a_{n-2}^{2}-4) $$
and so on, thus it can be easily proved by induction that
$$a_n^2-4=a_{n-1}^{2}a_{n-2}^{2}...a_2^2a_1^2(a_1^{2}-4) $$
thus
$$\sqrt{a_1^{2}-4}= \frac{\sqrt{a_n^2-4}}{a_1a_2...a_{n-2}a_{n-1}}\sim\frac{a_n}{a_1a_2...a_{n-2}a_{n-1}}$$
A: From $a_n^2-4 = a_1^2 \ldots a_{n-1}^2(a_1^2-4)$ one gets $\frac {a_n^2}{a_1^2 \ldots a_{n-1}^2}- \frac 4 {a_1^2 \ldots a_{n-1}^2}= (a_1^2-4)$ therefore $\frac {a_n}{a_1 \ldots a_{n-1}} = \sqrt {\frac 4 {a_1^2 \ldots a_{n-1}^2} + (a_1^2-4)} \tag 1$
Now, since $\lim a_n= +\infty$ it follows $\lim a_1^2 \ldots a_{n-1}^2 = +\infty$ therefore $\lim \frac {a_n}{a_1 \ldots a_{n-1}} = \sqrt {(a_1^2-4)}$
A: 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}
