<s>[INST] 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
 [/INST] 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
$$\sqrt{a_1^{2}-4}= \frac{\sqrt{a_n^2-4}}{a_1a_2...a_{n-2}a_{n-1}}\sim\frac{a_n}{a_1a_2...a_{n-2}a_{n-1}}$$
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