Came across this in a calc textbook from the 1800s and I can't figure out a way to solve it. Trying to write it in product form by taking the integral didn't work. I also tried adding consecutive values to see if it telescopes but I couldn't figure that out. Couldn't find any summation remotely similar on the web either.
Thanks for any help!
Note that $\cfrac{2}{1-x^{1/2^k}}-\cfrac{1}{1-x^{1/2^{k+1}}}=\cfrac{2-(1+x^{1/2^{k+1}})}{1-x^{1/2^k}}=\cfrac{1-x^{1/2^{k+1}}}{1-x^{1/2^k}}=\cfrac{1}{1+x^{1/2^{k+1}}}$. This means that your series telescopes i.e we have $$\sum_{k=0}^{\infty} \frac{1}{2^{k+1}(1+x^{1/2^{k+1}})}=\sum_{k=0}^{\infty} \left( \cfrac{1}{2^{k}(1-x^{1/2^k})}-\cfrac{1}{2^{k+1}(1-x^{1/2^{k+1}})} \right)=\cfrac{1}{1-x}-\cfrac{1}{\ln(x)}$$ Note this makes sense for $x\neq 1$. We also add in the extra term $n=0$ to obtain that $$\sum_{n=0}^{\infty} \cfrac{1}{2^n(1+\sqrt[2^n]{x})}=\cfrac{2}{1-x^2}-\frac{1}{\ln(x)}$$ for $x\neq 1, x>0$.