Where are the poles of the Riemann zeta function located?

Question

We know that $\zeta(s)$ has a simple pole at the point $s=1$ of residue $1$. Furthermore it can be seen here that the function $$ \eta(s)=\zeta(s)-\frac{1}{s-1} $$ is an entire function with no finite poles. On the other hand, we are told that the $\zeta$ function expressed as series is convergent for $\Re(s)>1$ and diverges for $\Re(s)<1$. What happens on the line $\Re(s)=1$? What is the behavior there? It can only be one of the two. My intuition tells me it's just one pole but I need someone to verify. Please help!

Answer 1

There is some confusion here. The series $\sum_{n=1}^\infty n^{-s}$ converges to $\zeta(s)$ when (and only when) $\operatorname{Re}(s)>1$. Nevertheless, the domain of the zeta function is $\Bbb C\setminus\{1\}$. It's similar to the situation of the function$$\begin{array}{rccc}f\colon&\Bbb C\setminus\{1\}&\longrightarrow&\Bbb C\\&z&\mapsto&\frac1{1-z}.\end{array}$$Its domain is $\Bbb C\setminus\{1\}$, but its Taylor series centered at $0$ is $\sum_{n=0}^\infty z^n$, which converges (to $f(z)$) when (and only when) $|z|<1$.

In particular, the only pole of $f$ is located at $1$.