Can someone help me finish this: evaluate $S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$ I am asked to find the closed form solution for the below.
$$S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$$
Just writing out the $S_1, S_2, S_3$, I have managed to find a pattern, which is:
$$S_n = \frac{S_{n-1}}{1-x^{2^n}} + \frac{x^{2^n}}{1-x^{2^{n+1}}}$$
I am not sure how to proceed onwards to solve this recurrence relation. Is there a clever trick I can do to solve it?

Just a kind of summary.

We have
\begin{align*}
S_n&=\sum_{j=1}^n\frac{x^{2^{j-1}}}{1-x^{2^{j}}}\\
&=\sum_{j=1}^n\left(\frac{x^{2^{j-1}}}{1-x^{2^{j-1}}}-\frac{x^{2^j}}{1-x^{2^j}}\right)\tag{1}\\
&\,\,\color{blue}{=\frac{x}{1-x}-\frac{x^{2^{n}}}{1-x^{2^{n}}}}\tag{2}
\end{align*}

Comment:


*

*In (1) we use the identity $a^2-b^2=(a-b)(a+b)$ with $a=1$ and $b=x^{2^{j-1}}$.

*In (2) we apply the telescoping series.
