Sum of the recursive series Let $\langle a_n \rangle$ be a recursive sequence given by $a_1>2$ and,
$a_{n+1}=a_n^2-2$  for $n \in \mathbb N$
Show that
$\sum_{n=1}^\infty \frac{1}{a_1a_2\cdots a_n} = \frac{a_1-\sqrt{a_1^2-4}}{2}$
I have reached this step:
$\frac{1}{a_1a_2a_3\cdots a_n}=\frac{1}{2} (\frac{a_n}{a_1a_2\cdots a_n-1} -\frac{a_n-1}{a_1a_2\cdots a_n})$
But I am not able to obtain the final expression .Please help me to obtain it. Thanks for help in advance.

Let $P_n = a_1 a_2 \cdots a_n$, and let $b_n = a_{n+1}/P_n$.
From the recursion, note that $$a_{n+1} - 2 = {a_n}^2 - 4 \implies a_n + 2 = \frac{a_{n+1} - 2}{a_n - 2},$$ hence we have the telescoping product $${P_n}^2 = \prod_{k=2}^{n+1} (a_k+2) = \prod_{k=2}^{n+1} \frac{a_{k+1} - 2}{a_k - 2} = \frac{a_{n+2} - 2}{a_2 - 2}$$
It follows that $${b_n}^2 = (a_2 - 2) \frac{{a_{n+1}}^2}{a_{n+2} - 2} = (a_2 - 2) \frac{a_{n+2} + 2}{a_{n+2} - 2} \implies \lim_{n\to\infty} b_n = \sqrt{a_2 - 2} = \sqrt{{a_1}^2 - 4}$$
Next, note that $$b_{k-1} - b_{k} =  \frac{{a_k}^2}{P_k} - \frac{a_{k+1}}{P_k}  = \frac{2}{P_n} \implies 2\sum_{k=2}^{n} \frac{1}{P_k} = b_1 - b_n $$ $$\implies 2\sum_{k=1}^{n} \frac{1}{P_k} = \frac{a_2+2}{a_1} - b_n = a_1 - b_n$$
We conclude that $$\sum_{k=1}^{\infty} \frac{1}{P_k} = \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{P_k} = \lim_{n\to\infty} \frac{a_1 - b_n}{2} = \frac{a_1 - \sqrt{{a_1}^2 - 4}}{2}$$
