$\zeta(s)$ Has Only One Pole In $\sigma>0$

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 $\zeta(s)=\sum1/n^{s}$ 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 $a\neq0$ 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.

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

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

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

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.

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

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 $\sigma>0$.