Trying to solve this question, it appears to be important to know the value of an infinite product
$$\displaystyle{\prod_{k=2}^\infty \left(1-\dfrac{1}{4^k}\right)}$$ which terms looks a lot like a "geometric series".
My question is: Is it possible to calculate the value of the product above?
More generally, Is there a general method to solve this kinds of infinite products? Is there a theory of infinite products that is more or less similar to the theory of infinite sums (i.e., series)? Where can I find it?
The convergence of an infinite product is surprisingly more easy than you think:
$$\prod_{k=1}^\infty\left(1-\frac1{4^k}\right)\text{ converges iff }\log\left[\prod_{k=1}^\infty\left(1-\frac1{4^k}\right)\right]\text{ converges}$$
$$\log\left[\prod_{k=1}^\infty\left(1-\frac1{4^k}\right)\right]=\sum_{k=1}^\infty\log\left(1-\frac1{4^k}\right)$$
And from there, it becomes an infinite sum/series problem. By the ratio test:
$$\lim_{n\to\infty}\left|\frac{\log\left(1-\frac1{4^{n+1}}\right)}{\log\left(1-\frac1{4^n}\right)}\right|=\lim_{n\to\infty}\left|\frac{4^n\left(1-\frac1{4^n}\right)}{4^{n+1}\left(1-\frac1{4^{n+1}}\right)}\right|=\frac14$$
Thus, it converges.
Nicely, according to WolframAlpha, we have
$$\sum_{k=1}^\infty\log\left(1-\frac1{4^{n+1}}\right)\approx-0.3731854421599476447\dots$$
And so
$$\prod_{k=1}^\infty\left(1-\frac1{4^k}\right)=e^{-0.3731854421599476447\dots}=0.68853753711614810750717\dots$$
Of course, these are approximations, not closed forms, but approximations satisfy the problem of "calculating" the product.