Infinite product: $(1-0.5^2)(1-0.5^3)(1-0.5^4)...$ Find a closed form for the value of the infinite product $(1-0.5^2)(1-0.5^3)(1-0.5^4)...$
I know it converges. At first I thought it was the Euler–Mascheroni constant, but it's only accurate to about 3 sig fig. I haven't been able to to solve it yet.
 A: $$A_n=\prod_{i=2}^n(1-(\frac{1}{2})^{k})=2 \left(\frac{1}{2};\frac{1}{2}\right)_n$$ where appears the Pochhammer symbol. For $$A_{\infty}=\prod_{i=2}^{\infty}(1-(\frac{1}{2})^{k})=2 \left(\frac{1}{2};\frac{1}{2}\right)_{\infty }\approx 0.5775761901732048425577994$$ while $$\gamma \approx 0.5772156649015328606065121$$ If now, you use what fixedp suggested and consider the sum of the logarithms and expand each of them as a Taylor series, you should find for the first terms $$\log(A_{\infty})=-x^2-x^3-\frac{3 x^4}{2}-x^5-\frac{11 x^6}{6}-x^7-\frac{7 x^8}{4}-\frac{4
   x^9}{3}-\frac{17 x^{10}}{10}-x^{11}-\frac{9 x^{12}}{4}-x^{13}-\frac{23
   x^{14}}{14}-\frac{23 x^{15}}{15}-\frac{15 x^{16}}{8}-x^{17}-\frac{19
   x^{18}}{9}-x^{19}-\frac{41 x^{20}}{20}+O\left(x^{21}\right)$$ in which you need to replace $x$ by $\frac{1}{2}$; this gives you $$\log(A_{\infty}) \approx -\frac{145045531}{264241152}$$ $$A_{\infty}\approx 0.5775770170415330689637466$$
Adding more terms leads quite quickly to the value.
