Sum of series $\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$ 
The problem is to find the sum of series $$\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$$
  to infinite terms, if $|x|<1$ 

I tried taking $\frac{x}{1-x^2}$ outside but could not solve it.
How to proceed further?

$${x\over1-x^2}+{x^2\over1-x^4}+{x^4\over1-x^8}+\cdots+{x^{2^n}\over1-x^{2^{n+1}}}=\left({1\over1-x}-{1\over1-x^2}\right)+\left({1\over1-x^2}-{1\over1-x^4}\right)+\left({1\over1-x^4}-{1\over1-x^8}\right)+\cdots+\left({1\over1-x^{2^n}}-{1\over1-x^{2^{n+1}}}\right)\\={1\over1-x}-{1\over1-x^{2^{n+1}}}\to{1\over1-x}-1={x\over1-x}$$
(Ah, I see achille hui posted the same answer while I was composing this.)
