<s>[INST] 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?
 [/INST] 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 $$\frac{x}{1-x}$$
</s>