<s>[INST] 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!
 [/INST] $$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}
$$
$$
\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}.(1-x^{2^{n-1}})}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}-x^{2^{n}}}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}\frac{-(1-x^{2^{n-1}})+1-x^{2^{n}}}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}(\frac{1}{1-x^{2^{n-1}}}-\frac{1}{1-x^{2^n}})
$$
Now consider upto $k$ terms then this can be written as 
(second term of $nth$ expression cancelled by first term of $(n+1)th$ expression) ,
$$
\frac{1}{1-x}-\frac{1}{1-x^{2^k}}
$$
For $k\to\infty$ and $x\in(0,1)$, 
$$
\frac{1}{1-x}-1
$$
Which is equal to 
$$
\frac{x}{1-x}
$$
</s>