For $x\neq -1$ and $x\neq1$, find
$$\lim_{n\to\infty} \sum_{k=1}^n \frac {x^{2^{k-1}}}{1-x^{2^k}}.$$
My attempts: I know that
$$\frac {x^{2^{k-1}}}{1-x^{2^k}} = \frac {1}{1-x^{2^k-1}} -\frac{1}{1-x^{2^k}}$$ for $k = 1, 2, 3, \cdots, n$. After that I am not able to proceed further…
Could you give me any hints?
Thanks in advance.
I presume the condition is $|z|\lt 1$, because otherwise the sum is unbounded.
Denote $z:=x^{2^{k-1}}$. Then, the expression inside the sum is $$\frac{z}{1-z^2}=\frac 12\frac{(1+z)-(1-z)}{(1-z)(1+z)}=\frac 12\left(\frac 1{1-z}-\frac 1{1+z}\right)=\sum_{r=0}^\infty z^{2r+1}$$
The required limit is,
$$\lim_{n\to\infty}\sum_{k=1}^n\sum_{r=0}^\infty z^{2r+1}=\lim_{n\to\infty}\sum_{k=1}^n\sum_{r=0}^\infty x^{2^{k-1}(2r+1)}\overset{\textrm{why?}}{=}\sum_{j=1}^\infty x^j=\ldots$$