Questions on convergence of nested series Say I have $\sum\limits_{n=1}^\infty e^{zn}$. We know that for $z \in \mathbb{C}$ and $n \in \mathbb{N}$ the series form of $e^{zn}$ converges. Does $\sum\limits_{n=1}^\infty e^{zn}$ converge then? Also for a series like $\sum\limits_{n=1}^\infty 1/n^z$ do we have convergence for all |z|>1 like in normal p-series? I guess what is really throwing me off in the second question is that I don't really know how to handle the complex exponential.
 A: Since
$$
|e^z|=|e^{\Re z+i\Im z}|=|e^{\Re z}e^{i\Im z}|=e^{\Re z} \quad \forall z\in \mathbb{C},
$$
the series $\sum_{n=1}^\infty e^{nz}$ converges for all $z\in \mathbb{C}$ with $e^{\Re z}<1$, i.e. for $\Re z<0$, and the series diverges for $\Re z\ge 0$. Furthermore
$$
\sum_{n=1}^\infty e^{nz}=\frac{e^z}{1-e^{z}} \quad \forall z \in \mathbb{C} \mbox{ with } \Re z<0.
$$
For all $z\in \mathbb{C}$ and $n\in \mathbb{N}$ we have
$$
\left|\frac{1}{n^z}\right|=\left|\frac{1}{n^{\Re z}}\frac{1}{n^{i\Im z}}\right|=\frac{|n^{-i\Im z}|}{n^{\Re z}}=\frac{1}{n^{\Re z}}
$$
Therefore the series $\sum_{n=1}^\infty 1/n^z$ converges for all $z\in \mathbb{C}$ with $\Re z>1$ and diverges for all $z\in \mathbb{C}$ with $\Re z\le 1$.
A: It is correct that this proves absolute convergence (which implies convergence) for this series if Rz>1, but the fact that the series does not converge absolutely does not imply divergence for Rz<=1.  Note that the terms of the alternating p-series have the same absolute values as the p-series, so the argument above would imply that the alternating p-series diverges for Rz<=1, but it converges (conditionally) for Rz in (0,1]
