sum of series $\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$ 
Sum of $n$ terms of the series
$$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots \cdots$$

Plan
$$\frac{x}{1-x^2}=\frac{1}{2}\frac{2x}{1-x^2}=\frac{1}{2}\bigg[\frac{1}{1-x}-\frac{1}{1+x}\bigg]$$
$$\frac{x^2}{1-x^4}=\frac{1}{2}\frac{2x^2}{1-x^4}=\frac{1}{2}\bigg[\frac{1}{1-x^2}-\frac{1}{1+x^2}\bigg]$$
Did not get any pattern to convert into Telescopic sum
How do i solve it Help me please

we can say that:
$$a_n=\frac{x^{2^n}}{1-x^{2^{n+1}}}$$
and we want to calculate:
$$S=\sum_{n=0}^\infty a_n$$
1 approach would be to try and use generating functions from the fact that:
$$S_{n+1}=S_n+\frac{x^{2^{n+1}}}{1-x^{2^{n+2}}}$$
Now you just need to evaluate this for whatever value of $n$ you need
