Limit of $(1-\frac{1}{2})*(1-\frac{1}{4})*\cdot\cdot\cdot*(1-\frac{1}{2^{n}})$? Using Monotone convergence theorem, we know the sequence ${x_{n}} = (1-\frac{1}{2})*(1-\frac{1}{4})*\cdot\cdot\cdot*(1-\frac{1}{2^{n}})$ has limit, but I can not derive the exactly limit, so how to calculate the limit of $x_{n}$?

Write
$$P=\prod_{k=1}^\infty \left(1-\frac{1}{2^n}\right)=\prod_{k=1}^4 \left(1-\frac{1}{2^n}\right)\prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right)=\frac{315}{1024}\prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right)$$ Now
$$\log\left(\prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right) \right)=\sum_{k=5}^\infty \log\left(1-\frac{1}{2^n}\right)\sim -\sum_{k=5}^\infty \frac{1}{2^n}=-\frac 1 {16}$$
$$\log\left(\prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right) \right)\sim -\frac 1 {16}\implies \prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right)\sim e^{-1/16}$$ So
$$P \sim \frac{315}{1024}e^{-1/16} =0.288980\cdots$$ while the "exact" value is $0.288788\cdots$
Edit
Using q-Pochhammer symbols, the formal expression of the infinite product
$$P=\prod_{n=1}^\infty \left(1-\frac{1}{2^n}\right)=\left(\frac{1}{2};\frac{1}{2}\right)_{\infty }$$ If you look at equation $(9)$ in this page, you will find an approximation which, for your case, write
$$P \sim 2^{13/24} \sqrt{\frac{\pi }{\log (2)}} \exp\Bigg[-\frac{\pi ^2}{6 \log (2)} \Bigg]$$ which is a a relative error of $2.22\times 10^{-14}$.
