If a series converges then the power series converges for all z

Question

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 $|z|\neq1$?

Answer 1

Some hints:

• 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 $(1)$.

Answer 2

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 $n \rightarrow \infty$.