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!

As soon as $|x|<1$ we have
$$ \frac{x^{2^{n-1}}}{1-x^{2^n}} = x^{2^{n-1}}+x^{3\cdot 2^{n-1}}+x^{5\cdot 2^{n-1}}+\ldots \tag{1}$$
hence:
$$ \sum_{n\geq 1}\frac{x^{2^{n-1}}}{1-x^{2^n}} = \sum_{m\geq 0}\sum_{h\geq 0} x^{(2h+1) 2^m} = \sum_{n\geq 1} x^n = \frac{x}{1-x}\tag{2} $$
since every positive integer can be represented in a unique way as the product between a power of two and an odd integer.
