If $a_{k}=2^{2^k}+2^{-2^k}$ then evaluate $\prod_{k=1}^\infty\left(1-\frac{1}{a_{k}}\right)$ If $$a_{k}=2^{2^k}+2^{-2^k}$$ then evaluate $$\prod_{k=1}^\infty\left(1-\frac{1}{a_{k}}\right)$$
I tried using Sophie-Germaine Identity about factorisation for $x^4+4$ but it did not work
 A: Let $2^{2^{-k}}=b, 2^{2^k}=\dfrac1b $
$1-\dfrac1{a_k}=\dfrac{b^2-b+1}{b^2+1}$
$a_{k+1}=(2^{2^k})^2+(2^{2^{-k}})^2=b^2+\dfrac1{b^2}=\dfrac{b^4+1}{b^2}$
$1-\dfrac1{a_{1+k}}=\dfrac{b^4-b^2+1}{b^4+1}$
Observe that $$(b^2-b+1)(b^2+b+1)=(b^2+1)^2-b^2=b^4+b^4+1$$
and $$(1+b^2)(1-b^2)=1-b^4$$
$$\implies\prod_{m=k}^n\left(1-\dfrac1{a_m}\right)=\dfrac{1-b^2}{1+b+b^2}\cdot\dfrac{1-b^{2^n}+b^{2^{n+1}}}{1-b^{2^{n+1}}}$$  as for $1-b^2\ne0$
Observe that $n\to\infty, b^{2^n}=0$
Here $k=1,4b=1$
A: Note that 
$$1 - \frac{1}{a_k} = \frac{1 - 2^{-2^k} + 4^{-2^k}}{1 + 4^{-2^k}} = \frac{1 + 8^{-2^k}}{(1 + 2^{-2^k})(1 + 4^{-2^k})}$$
and for $|x| < 1$, we have 
$$\prod_{k=1}^{\infty} (1 + x^{2^k}) = \lim_{n \to \infty} \prod_{k=1}^n (1 + x^{2^k}) = \lim_{n \to \infty} \frac{1 - x^{2^{n+1}}}{1 - x^2} = \frac{1}{1-x^2}$$
which gives
$$\prod_{k=1}^\infty \left(1 - \frac{1}{a_k}\right) = \frac{\prod_{k=1}^\infty (1 + 8^{-2^k})}{\prod_{k=1}^\infty (1 + 2^{-2^k})\prod_{k=1}^\infty (1 + 4^{-2^k})} = \frac{\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{16}\right)}{1 - \frac{1}{64}} = 
 \frac{5}{7}.$$
