Evaluate series $\sum\limits_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$

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.

$$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}} $$ $$ \sum_{n=1}^{\infty}\frac{x^{2^{n-1}}.(1-x^{2^{n-1}})}{(1-x^{2^n}).(1-x^{2^{n-1}})} $$ $$ \sum_{n=1}^{\infty}\frac{x^{2^{n-1}}-x^{2^{n}}}{(1-x^{2^n}).(1-x^{2^{n-1}})} $$ $$ \sum_{n=1}^{\infty}\frac{-(1-x^{2^{n-1}})+1-x^{2^{n}}}{(1-x^{2^n}).(1-x^{2^{n-1}})} $$ $$ \sum_{n=1}^{\infty}(\frac{1}{1-x^{2^{n-1}}}-\frac{1}{1-x^{2^n}}) $$ 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 $$ \frac{x}{1-x} $$

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