Find the partial sum formula of $\sum_{i=1}^n \frac{x^{2^{i-1}}}{1-x^{2^i}}$

Question

Given next series:

$$\frac{x}{1 - x^2} + \frac{x^2}{1 - x^4} + \frac{x^4}{1 - x^8} + \frac{x^8}{1 - x^{16}} + \frac{x^{16}}{1 - x^{32}} + ... $$

and $|x| < 1$. Need to derive $S_n$ formula from series partial sums.

I could only find that $S_{k+1}=\frac{S_k}{1-x^{2^k}} + \frac{x^{2^k}}{1-x^{2^{k+1}}}$. But this is incorrect answer, of course, but I don't know what to do next...

Thank you in advance!

Answer 1

By induction you can proof easily $$\sum\limits_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^k}} = \frac{1}{1-x^{2^n}}\sum\limits_{k=1}^{2^n-1}x^k$$ and with $\enspace\displaystyle \sum\limits_{k=1}^{2^n-1}x^k = \frac{x-x^{2^n}}{1-x}\enspace$ the formula is complete.

Answer 2

Hint: Use method of cancellation $$\frac {x^2} {1-x^4} = \left[\frac 1 {1-x^2} - \frac 1 {1-x^4}\right]$$ $$\frac {x^4} {1-x^8} = \left[\frac 1 {1-x^4} - \frac 1 {1-x^8}\right]$$