Given the following infinite product: $$\lim_{n\to \infty} \frac{1}{2}\frac{3}{4}\cdots \frac{2^n-1}{2^n}.$$
It is easy to see that above infinite product is convergent by the convergence of the following series: $$\sum_{n=1}^{\infty}\ln \left(1-\frac{1}{2^n}\right)$$.
But I do not know how to calculate the infinite product. I would appreciate it if someone can give any suggestions and comments.
For the infinite product $$P=\frac{\prod_{i=0}^\infty (2^i-1) } {\prod_{i=0}^\infty 2^i }=\frac{315}{1024}\frac{\prod_{i=5}^\infty (2^i-1) } {\prod_{i=5}^\infty 2^i }$$
$$A=\log \left(\frac{\prod_{i=5}^\infty (2^i-1) } {\prod_{i=5}^\infty 2^i } \right)=\sum_{i=5}^\infty\log \left(1-\frac1 {2^i}\right)=-\sum_{i=5}^\infty \Big[\frac1 {2^i}+ \frac1 {2^{2i}}+\frac1 {2^{3i}}+\cdots\Big]$$ and we face geometric progressions. So, summing up to infinty, we have $$A=-\Big[\frac{1}{16}+\frac{1}{768} +\frac{1}{28672}+\frac{1}{983040}+\frac{1}{32505856}+\frac{1}{1056964608}+\cdots\Big]$$ Using only these numbers $$A \sim -\frac{10458550091}{163829514240}$$ Now using the approximation $$e^A=\frac{1+\frac{A}{2}+\frac{A^2}{12}}{1-\frac{A}{2}+\frac{A^2}{12}}=\frac{311910183016998882664441}{332471213189757736214521}$$ $$P=\frac{98251707650354648039298915}{340450522306311921883669504}$$ which, converted to decimals is $P=0.28859$ while the "exact" value is $P=0.28879$
You maybe could simplify the denominator of the infinite product since $\prod_{n=1}^{m}2^n=2^{\sum_{n=1}^{m}n}=2^{\frac{m(m+1)}{2}}$ which would mean that $\lim_{m \rightarrow \infty} \prod_{n=1}^{m}\frac{2^n-1}{2^n}=\lim_{m \rightarrow \infty}\frac{1}{2^{\frac{m(m+1)}{2}}} \prod_{n=1}^{m}(2^n-1)$
