What is the value of $\prod_{i=1}^\infty \left(1-\frac{1}{2^i}\right)$? [duplicate]

$$\sum_{k\geq 1}\log\left(1-\frac{1}{2^k}\right) = -\sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{n 2^{kn}}=-\sum_{m\geq 1}\frac{1}{2^m}\sum_{d\mid m}\frac{1}{d}=-\sum_{m\geq 1}\frac{\sigma_1(m)}{m 2^m}$$ hence: $$\prod_{i\geq 1}\left(1-\frac{1}{2^i}\right)=\exp\left(-\sum_{m\geq 1}\frac{\sigma_1(m)}{m 2^m}\right)\geq\exp\left(-\sum_{m\geq 1}\frac{m+1}{2^{m+1}}\right)=e^{-3/2}$$ but the bound: $$\sigma_1(m)=\sum_{d\mid m}d \leq \sum_{k=1}^{m}k = \frac{m(m+1)}{2}$$ is obviously very crude.

You can use the $q$-Pochhammer symbol to represent the infinite product as

$$ \prod_{i=1}^{\infty}\left( 1-\frac{1}{2^i} \right) = \left( \frac{1}{2}, \frac{1}{2} \right)_{\infty} . $$

Note:

$$ \left( a, q \right)_{\infty}= \prod_{i=1}^{\infty}\left( 1-a q^k \right) $$