<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] 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$$\left|\frac x{1-x}-\sum_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^x}}\right|<\varepsilon.$$
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