Solve Recurrence Relation $a_k=(a_{k-1})^2-2$ $$a_k=\left(a_{k-1}\right)^2-2$$
$a_0=\frac{5}{2}$
Then find $$P=\prod_{k=0}^{\infty} \left(1-\frac{1}{a_k}\right)$$
My try:
I rewrote the Recurrence equation as
$$a_k+1=(a_{k-1}-1)(a_{k-1}+1)$$ $\implies$
$$\frac{1}{a_{k-1}-1}=\frac{a_{k-1}+1}{a_k+1}$$ $\implies$
$$\frac{a_{k-1}}{a_{k-1}-1}=\frac{(a_{k-1})^2+a_{k-1}}{a_k+1}=\frac{a_k+a_{k-1}+2}{a_k+1}$$
any hint here?

Hint: There are three steps.  First, for each $k=0,1,2,\ldots$, show that $$a_k=2^{2^k}+\frac{1}{2^{2^k}}\,.$$
Second, write $$1-\frac{1}{a_k}=\frac{a_k-1}{a_k}=\left(\frac{a_{k+1}+1}{a_k+1}\right)\frac{1}{a_k}\,,$$
for all $k=0,1,2,\ldots$.
Finally, show that
$$\prod_{k=0}^n\,a_k=\frac{2}{3}\left(2^{2^{n+1}}-\frac{1}{2^{2^{n+1}}}\right)\,,$$ using the identity
$$(x-y)\,\prod_{k=0}^n\,\left(x^{2^k}+y^{2^k}\right)=x^{2^{n+1}}-y^{2^{n+1}}\,,$$
for all $n=0,1,2,\ldots$.

 You should in the end obtain $$\prod_{k=0}^\infty\,\left(1-\frac{1}{a_k}\right)=\lim_{n\to\infty}\,\frac{3}{7}\left(\frac{2^{2^{n+1}}+1+\frac{1}{2^{2^{n+1}}}} { 2^{2^{n+1}}-\frac{1}{2^{2^{n+1}}} }\right)=\frac{3}{7}\,.$$

