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!
 A: 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.
A: $$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}
$$
$$
\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}.(1-x^{2^{n-1}})}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}-x^{2^{n}}}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}\frac{-(1-x^{2^{n-1}})+1-x^{2^{n}}}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}(\frac{1}{1-x^{2^{n-1}}}-\frac{1}{1-x^{2^n}})
$$
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 
$$
\frac{x}{1-x}
$$
A: Let $x\in(0,1)$ and let $\varepsilon>0$. I shall prove that there is a natural number $p$ such that$$n\geqslant p\Longrightarrow\left|\frac x{1-x}-\sum_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^x}}\right|<\varepsilon.$$Take $p'\in\mathbb N$ such that $\left|\frac x{1-x}-(x+x^2+\cdots+x^{p'})\right|<\varepsilon$. Take $p\in\mathbb N$ such that, when you express any element of $\{1,2,\ldots,p'\}$ as the product of an odd number with a power of $2$, then the exponent of that power of $2$ is always smaller than or equal to $p$. Then $$n\geqslant p\Longrightarrow x+x^2+\cdots+x^{p'}\leqslant\sum_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^x}}<x+x^2+x^3+\cdots=\frac x{1-x}$$and therefore$$\left|\frac x{1-x}-\sum_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^x}}\right|<\varepsilon.$$
A: Adding term by term: 
$$S=\frac{x(1+x^2)+x^2}{1-x^4}+\frac{x^2}{1-x^8}+...=$$
$$\frac{(x+x^2+x^3)(1+x^4)+x^4}{1-x^8}+\frac{x^8}{1-x^{16}}+...=$$
$$\frac{\sum_{n=1}^{\infty} x^n}{1-x^{2\cdot \infty}}=$$
$$\frac{x}{1-x}$$
because $0<x<1$.
