Does anyone know how to calculate the value of the sum
$$\sum_{n=0}^\infty \frac{1}{x^{2^n}+1}=\text{?}$$
for any values of $x$ for which it converges? I know how to evaluate some similar sums, such as $$\sum_{n=0}^\infty \frac{2^n}{x^{2^n}+1}=\frac{1}{x-1}$$ and $$\sum_{n=0}^\infty \frac{2^{-n}}{x^{2^{-n}}+1}=\frac{2}{1-x^2}+\frac{1}{\ln(x)}$$ Both of which telescope. Is there any way to make the first sum telescope (or calculate it by other methods)? Interestingly, the reason I want to calculate it is because it is related to the generating function of the base 2 "digit sum" sequence: $$\sum_{n=0}^\infty \frac{s_2(n)}{x^n}=\frac{x}{x-1}\sum_{n=0}^\infty \frac{1}{x^{2^n}+1}$$
Assuming $x>1$, $$ \frac{1}{1+x^{2^n}} = x^{-2^n}-x^{-2\cdot 2^n}+x^{-3\cdot 2^n}-\ldots $$ Every $m\in\mathbb{N}^*$ can be written in a unique way as $(2k+1)2^{h}$, hence by summing both sides of the previous line on $n\geq 0$ we get $$ \sum_{n\geq 0}\frac{1}{1+x^{2^n}} = \sum_{m\geq 1}\frac{1-\nu_2(m)}{x^m}=\sum_{m=2k+1}\frac{1}{x^m}+\sum_{m=4k+1}\frac{0}{x^m}+\sum_{m=8k+4}\frac{-1}{x^m}+\ldots$$ or $$ \frac{x}{x^2-1}-\frac{x^4}{x^8-1}-\frac{2x^8}{x^{16}-1}-\frac{3x^{16}}{x^{32}-1}-\ldots $$ where the residue at $x=1$ equals $$ \frac{1}{2}-\sum_{u\geq 1}\frac{u}{2^{u+2}}=0 $$ as it should. Not sure this can be turned into something with a "nice" closed form.