Evaluate series $\sum\limits_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$ 
Determine the value of
  $$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$$
  or $$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$$
  for $x\in\mathbb{R}$.

The answer is $\dfrac{x}{1-x}$ for $x\in(0,1)$. To prove this, notice
$$\frac{x}{1-x^2}=x+x^3+x^5+\cdots$$
$$\frac{x^2}{1-x^4}=x^2+x^6+x^{10}+\cdots$$
$$\cdots$$
Add them all and get the answer. Unfortunately, I havn't got a direct method to calculate it. Appreciate for your help!

Adding term by term: 
$$S=\frac{x(1+x^2)+x^2}{1-x^4}+\frac{x^2}{1-x^8}+...=$$
$$\frac{(x+x^2+x^3)(1+x^4)+x^4}{1-x^8}+\frac{x^8}{1-x^{16}}+...=$$
$$\frac{\sum_{n=1}^{\infty} x^n}{1-x^{2\cdot \infty}}=$$
$$\frac{x}{1-x}$$
because $0<x<1$.
