If a series converges then the power series converges for all z How can I prove that if $\sum \limits_{n=1}^{\infty} c_n$ , $c_n\in \mathbb{C}$, converges then $\sum \limits_{n=1}^{\infty} c_n  \frac{z^n}{1-z^n}$ converges for all z in $\mathbb{C}$ with $|z|\neq1$?
 A: Some hints:


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*When $\sum_{k=1}^\infty c_k$ converges then $\sum_{k=1}^\infty c_k\>\rho^k$ converges absolutely for every positive $\rho<1$.

*Distinguish the cases $|z|<1$ (easier case) and $|z|>1$.


Solution:
When $\sum_{k=1}^\infty c_k$ converges then there is an $M>0$ with $|c_k|\leq M$ for all $k\geq1$.
Assume $|z|=:\rho<1$. As $|z^k|=\rho^k\leq\rho$ for all $k\geq1$ the series
$$\sum_{k=1}^\infty {M\over1-\rho}\>\rho^k$$
is a convergent majorant of $S:=\sum_{k=1}^\infty{z^k\over 1-z^k}$. It follows that $S$ converges absolutely.
Assume $|z|=:{1\over\rho}>1$, which implies that $0<\rho<1$. Then
$${z^k\over 1-z^k}=-1+{1\over 1-z^k}=-1+{z^{-k}\over z^{-k}-1}\ .$$
It follows that
$$\sum_{k=1}^\infty c_k{z^k\over1-z^k}=-\sum_{k=1}^\infty c_k +\sum_{k=1}^\infty c_k{z^{-k}\over z^{-k}-1}\ .\tag{1}$$
As $|z^{-k}|=\rho^k\leq\rho$ for all $k\geq1$ the series
$$\sum_{k=1}^\infty {M\over1-\rho}\>\rho^k$$
is a convergent majorant of the second series appearing on the right of $(1)$.
A: It suffices to show that for any $z$ not on the unit circle, there is a natural number $C_z$ so that $|\frac{z^n}{1-z^n}|<C_z$ for all $n \in N$. Given this, $|\frac{z^n}{1-z^n}c_n| \leq C_z\cdot|c_n|$ for each $n$, and absolute convergence of $\sum_{n=1}^\infty c_n$ implies that of $\sum_{k=1}^\infty\frac{z^n}{1-z^n}c_n$. 
To prove that $C_z$ exists for $|z| \neq 1$, you might try to show $\frac{z^n}{1-z^n}$ has a limit as $n \rightarrow \infty$.
