Finding the minimum value of this product Is it true that $f(n) = \prod _{ i=1 }^{ n }{ (1-\frac { 1 }{ { 2 }^{ i } } ) } \ge \frac{1}{4} \quad \forall n$? I came up with this expression while trying to find an alternative way to solve a physics problem. The official solution has a minimum value of $\frac{1}{4}$, but I couldn't prove that my solution also has a minimum of $\frac{1}{4}$.

For $0 < x \leqslant \frac12$, we have the estimate
$$\lvert \log (1-x)\rvert = \sum_{n=1}^\infty \frac{x^n}{n} < x + \frac{x^2}{2}\sum_{k=0}^\infty x^k = x + \frac{x^2}{2(1-x)} \leqslant x + x^2,$$
so we find
$$\log f(n) = \sum_{i=1}^n \log \left(1 - \frac{1}{2^i}\right) > - \sum_{i=1}^n \left(\frac{1}{2^i} + \frac{1}{2^{2i}}\right) > - \left(1 + \frac13\right),$$
whence
$$f(n) > e^{-4/3} \approx 0.26359713811572677 > \frac14.$$
A less crude estimate of the logarithm yields larger bounds.
