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

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

Answer 1

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}

Answer 2

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

Answer 3

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

$$\sqrt{a_1^{2}-4}= \frac{\sqrt{a_n^2-4}}{a_1a_2...a_{n-2}a_{n-1}}\sim\frac{a_n}{a_1a_2...a_{n-2}a_{n-1}}$$