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

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...

• What about $$S_{k+1}=S_k+\frac{x^{2^k}}{1-x^{2^{k+1}}}$$ – Dr. Mathva Apr 15 at 8:58
• Yes, probably. But my problem is I don't know how to get the sum of series $S_n$ from this. I think I need to get $S_k$ without dependencies from $S_{k+1}$ and in finale formula of $S_n$ there should be no sums or something. – Alex Apr 15 at 9:04

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.
Hint: Use method of cancellation $$\frac {x^2} {1-x^4} = \left[\frac 1 {1-x^2} - \frac 1 {1-x^4}\right]$$ $$\frac {x^4} {1-x^8} = \left[\frac 1 {1-x^4} - \frac 1 {1-x^8}\right]$$
• Thank you! I think this is a kind of geometric progression, right? But I can't derive $q$ and as follows the $S_n$ from the series, and don't understand how your hint is applicable to do this. Can you give me some more hints maybe? :) Again, thank you! – Alex Apr 15 at 9:23