ζ(s) Has Only One Pole In σ>0

Here we show that the Riemann Zeta function has only one pole in the domain $\sigma >0$. It is based on a suggestion by user @leoli1 on math.stackexchange.

The video for this blog is here [youtube], and the slides are here [pdf].

Previously

The Riemann Zeta function represented by the series $\zeta (s)=\sum 1/n^s$ converges for $\sigma >1$, and therefore has no poles in that domain.

In the last blog post developed a new series for $\zeta (s)$ based on the eta function $\eta (s)$.

$$\zeta (s)=\frac{1}{1-2^{1-s}}\eta (s)$$

Because $\eta (s)$ converges for $\sigma >0$ (Dirichlet series), any divergence must come from the factor $(1-2^{1-s}{)}^{-1}$. 

The denominator $(1-2^{1-s})$ is zero at $s=1+2\pi ia/\ln (2)$ for integer $a$, so the factor $(1-2^{1-s}{)}^{-1}$ diverges at all these points. 

Visualising $\zeta (s)$ suggested it had only one pole at $s=1+0i$. If true, this would require $\eta (s)$ to have zeros at $s=1+2\pi ia/\ln (2)$ for integers $a\ne 0$ to cancel out the other poles from $(1-2^{1-s}{)}^{-1}$. 

To prove this directly isn't easy, but there is a nice indirect path.

Yet Another Series for $\zeta (s)$

We start with a specially constructed Dirichlet series.

$$X(s)=\frac{1}{1^s}+\frac{1}{2^s}-\frac{2}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}-\frac{2}{6^s}+\dots$$

The pattern can be exploited to find yet another series for $\zeta (s)$.

$$\begin{array}{rl}\zeta (s)-X(s)& =\frac{3}{3^s}+\frac{3}{6^s}+\frac{3}{9^s}+\dots \\ \\ & =\frac{3}{3^s}\zeta (s)\\ \\ \zeta (s)& =\frac{1}{1-3^{1-s}}X(s)\end{array}$$

Comparing Potential Poles

The Dirichlet series $X(s)$ converges for $\sigma >0$ (Dirichlet series), so any divergence must come from the factor $(1-3^{1-s}{)}^{-1}$. The denominator $(1-3^{1-s})$ is zero when $s=1+2\pi ib/\ln (3)$ for integer $b$.

We can equate the two expressions for where the poles of $\zeta (s)$ could be.

$$\begin{array}{rl}1+\frac{2\pi ia}{\ln (2)}& =1+\frac{2\pi ib}{\ln (3)}\\ \\ \frac{a}{b}& =\frac{\ln (2)}{\ln (3)}\end{array}$$

There are no non-zero integers $a$ and $b$ which satisfy this because $\ln (2)/\ln (3)$ is irrational.

This leaves us with $s=1+0i$ as the only pole for $\zeta (s)$ in the domain $\sigma >0$.