<s>[INST] 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
 [/INST] 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
$$\prod_{k=1}^\infty \left(1 - \frac{1}{a_k}\right) = \frac{\prod_{k=1}^\infty (1 + 8^{-2^k})}{\prod_{k=1}^\infty (1 + 2^{-2^k})\prod_{k=1}^\infty (1 + 4^{-2^k})} = \frac{\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{16}\right)}{1 - \frac{1}{64}} = 
 \frac{5}{7}.$$
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