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

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 $\lim \frac {a_n}{a_1 \ldots a_{n-1}} = \sqrt {(a_1^2-4)}$
