$\{a_n\}$ is a recursive sequence given by setting $a_1\gt2, a_{n+1}=a_{n}^2-2, \forall n\in\mathbb N$. Then, show that $$\sum_{n=1}^\infty\frac{1}{a_1a_2\cdots a_n}=\frac{a_1-\sqrt{a_1^2-4}}{2}$$

Due to the recursive relation, we can re-write the given series as: $$\sum_{n=1}^\infty\frac{1}{a_1a_2\cdots a_n}={1\over2}\sum_{n=1}^\infty\frac{a_n^2-a_{n+1}}{a_1a_2\cdots a_n}={1\over2}\lim_{m\to\infty}\sum_{n=1}^m\left(\frac{a_n}{a_1a_2\cdots a_{n-1}}-\frac{a_{n+1}}{a_1a_2\cdots a_n}\right)\\={1\over2}\lim_{m\to\infty}\left({a_1\over1}-\frac{a_{m+1}}{a_1a_2\cdots a_m}\right)$$ Ultimately, I have to show that $$\lim_{m\to\infty}\left(\frac{a_{m+1}}{a_1a_2\cdots a_m}\right)=\sqrt{a_1^2-4}=\sqrt{a_2-2}$$ First of all, since $a_1>2$, it is not difficult to show that $a_{n+1}=a_n^2-2>a_n$ inductically. Also, I knew that $\{a_n\}$ is not bounded above by drawing a graph of $y=x$ and $y=x^2-2$. Consequently, I thought that the only way to show this is using squeeze. But, I'm actually lost the way since it is quite hard to make an inequality(in order to use the squeeze) since the limit converges to nonzero value. Could you please suggest the idea for having the value of the limits? Thanks.

  • 1
    $\begingroup$ The same idea as here works: math.stackexchange.com/questions/3561125/… $\endgroup$ – LHF Mar 2 at 13:53
  • 1
    $\begingroup$ But $a_1^2-4=a_2-2$ $\endgroup$ – Hagen von Eitzen Mar 2 at 13:55
  • $\begingroup$ @Atticus Oh, thanks sir!:D $\endgroup$ – ToBY Mar 2 at 13:57
  • $\begingroup$ @HagenvonEitzen It was typo and I edited. Thanks:) $\endgroup$ – ToBY Mar 2 at 13:57