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
 A: Statement:

$$S_n=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots+\frac{x^{2^{n-1}}}{1-x^{2^{n}}}=\frac{x-x^{2^n}}{(1-x)(1-x^{2^n})}\tag{1}$$

How did I invent formula (1)? By adding a few items by hand and looking into the pattern that started to emerge. Let us prove it by induction:
For $n=1$:
$$S_1=\frac{x-x^2}{(1-x)(1-x^2)}=\frac{x}{1-x^2}$$
So the statement is true for $n=1$. 
Now the induction step:
$$S_{n+1}=S_n+\frac{x^{2^{n}}}{1-x^{2^{n+1}}}=\frac{x-x^{2^n}}{(1-x)(1-x^{2^n})}+\frac{x^{2^{n}}}{1-x^{2^{n+1}}}$$
$$S_{n+1}=\frac{x-x^{2^n}}{(1-x)(1-x^{2^n})}+\frac{x^{2^{n}}}{(1-x^{2^{n}})(1+x^{2^{n}})}$$
$$S_{n+1}=\frac{(x-x^{2^n})(1+x^{2^n})+x^{2^{n}}(1-x)}{(1-x)(1-x^{2^n})(1+x^{2^n})}$$
$$S_{n+1}=\frac{x+x^{2^n+1}-x^{2^n}-x^{2^{n+1}}+x^{2^n}-x^{2^n+1}}{(1-x)(1-x^{2^n})(1+x^{2^n})}$$
$$S_{n+1}=\frac{x-x^{2^{n+1}}}{(1-x)(1-x^{2^{n+1}})}$$
Done.
A: 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
