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?
 A: If you combine the first two terms, you should end up with $$\frac{x(1+x^2)}{1-x^4}+\frac{x^2}{1-x^4}=\frac{x+x^2+x^3}{1-x^4}$$
If you combine the first three terms, you should end up with $$\frac{(x+x^2+x^3)(1+x^4)}{1-x^8}+\frac{x^4}{1-x^8}=\frac{x+x^2+x^3+x^4+x^5+x^6+x^7}{1-x^8}$$
Following this pattern, you will find out that the sum of the first $n$ terms is $$\frac{x}{1-x^{2^n}}\sum_{i=0}^{2^n-2}x^i$$
And you should be able to evaluate this sum, because it is the finite sum of a geometric progression. Taking the limit as $n\to\infty$ should yield your final answer of $$\frac{x}{1-x}$$
A: $${x\over1-x^2}+{x^2\over1-x^4}+{x^4\over1-x^8}+\cdots+{x^{2^n}\over1-x^{2^{n+1}}}=\left({1\over1-x}-{1\over1-x^2}\right)+\left({1\over1-x^2}-{1\over1-x^4}\right)+\left({1\over1-x^4}-{1\over1-x^8}\right)+\cdots+\left({1\over1-x^{2^n}}-{1\over1-x^{2^{n+1}}}\right)\\={1\over1-x}-{1\over1-x^{2^{n+1}}}\to{1\over1-x}-1={x\over1-x}$$
(Ah, I see achille hui posted the same answer while I was composing this.)
A: If $|x|<1$, for any $k\in\mathbb{N}$ we have
$$ \frac{x^{2^k}}{1-x^{2^{k+1}}} = x^{2^k}+x^{3\cdot 2^k}+x^{5\cdot 2^k}+\ldots\tag{A}$$
but given any $n\in\mathbb{N}^+$, there is a unique representation of $n$ as $2^k\cdot d$, with $k\in\mathbb{N}$ and $d\in 2\mathbb{N}+1$. It follows that:
$$ \sum_{k\geq 0}\frac{x^{2^k}}{1-x^{2^{k+1}}} = \sum_{n\geq 1} x^n = \color{red}{\frac{x}{1-x}}.\tag{B}$$
A: Notice $$\frac{y}{1-y^2} = \frac{(1+y)-1}{1-y^2} = \frac{1}{1-y} - \frac{1}{1-y^2}$$
The sum at hand is a telescoping one. The partial sums has the form:
$$\sum_{n=1}^p
\frac{x^{2^{n-1}}}{1-x^{2^n}} = 
\sum_{n=1}^p
\left[\frac{1}{1-x^{2^{n-1}}} - \frac{1}{1-x^{2^n}}\right]
= \frac{1}{1-x} - \frac{1}{1-x^{2^p}}$$
When $|x| < 1$, $x^{2^p} \to 0$. This leads to
$$\sum_{n=1}^\infty
\frac{x^{2^{n-1}}}{1-x^{2^n}}
 = 
\lim_{p\to\infty}
\left[\frac{1}{1-x} - \frac{1}{1-x^{2^{p}}}\right]
= \frac{1}{1-x} - \frac{1}{1-0} = \frac{x}{1-x}$$
