I was thinking a way to evaluate the product $$\prod_{n=1}^\infty \frac{1}{1-2^{-n}}=\prod_{n=1}^\infty \left(1+2^{-n}+2^{-2n}+\dots \right)$$
I just thought about the function $f(x)=\prod_{n=0}^\infty \left( 1+\frac{x}{2^n} \right)$. These functions of the form $\prod_{n=0}^\infty \left( 1+\frac{x}{a_n} \right)$ has an infinite series: $1+x\sum_{n=0}^\infty \frac{1}{a_n} +x^2\sum_{n=0}^\infty\sum_{k>n} \frac{1}{a_n a_k}+...$ Desenvolving this function with $a_n=2^n$ I got: $$f(x)=\prod_{n=0}^\infty \left( 1+\frac{x}{2^n} \right)=\sum_{n=0}^\infty \left( \frac{x}{2^{(n-1)/2}}\right)^n\prod_{k=1}^n \frac{1}{1-\frac{1}{2^k}}$$ This result is explained here. So, if we suppose there exists an $m \in \mathbb{N}$, then we have that $$\frac{f^{(m)}(x)}{m!}=\frac{1}{m!}\sum_{n\geq m}\frac{n!}{(n-m)!}\frac{x^{n-m}}{2^{n(n-1)/2}}\prod_{k=1}^n\frac{1}{1-\frac{1}{2^k}}$$ So, setting $x=0$, then we have that $$\frac{f^{(m)}(0)}{m!}=\frac{1}{2^{m(m-1)/2}}\prod_{k=1}^m \frac{1}{1-\frac{1}{2^k}}$$ Multiplying $2^{m(m-1)/2}$ both sides and taking $m\rightarrow \infty$, then we have that $$\prod_{n=1}^\infty \frac{1}{1-\frac{1}{2^n}}=\lim_{m\rightarrow \infty} \frac{2^\frac{m(m-1)}{2}}{m!}\frac{d^m}{dx^m}\left(\prod_{n=0}^\infty \left(1+\frac{x}{2^n}\right)\right)\vert_{x=0}$$
I have a approximation for that product, which is about 3.46274661945. Does this product has a closed form?
If someone can help me, I will be grateful.
I think a comment wouldn't have that space, so I'm answering as well.
The product $\prod_{n=1}^\infty (1+\frac{x}{2^n})=\frac{f(x)}{x+1}$. Since $f(-1)=0$, and if we let $x\rightarrow -1$, we will have that $$\prod_{n=1}^\infty (1-\frac{1}{2^n})=\lim_{x\rightarrow-1} \frac{f(x)-f(-1)}{x-(-1)}=f'(-1)$$ So: $$\prod_{n=1}^\infty \frac{1}{1-\frac{1}{2^n}}=\lim_{N\rightarrow \infty} \frac{2^{N(N-1)/2}f^{(N)}(0)}{N!}=\frac{1}{f'(-1)}$$
Since $f(x)=1+\frac{1}{2}x+\frac{4}{3}x^2+...$, so $f'(x)=\frac{1}{2}+\frac{8}{3}x+...+n\frac{x^{n-1}}{2^{n(n-1)/2}}\prod_{k=1}^n \frac{1}{1-\frac{1}{2^n}}+\dots$, and we have that: $$\prod_{n=1}^\infty \frac{1}{1-\frac{1}{2^n}}=\frac{1}{\frac{1}{2}-\frac{8}{3}+\frac{8}{7}-\dots}$$ A little approach that I have viewing the OEIS link that @TheSilverDoe commented.