Trivial Zeroes of Riemann Zeta Function are Even Negative Integers

Theorem
Suppose $\rho$ is a zero of the Riemann zeta function not contained in the critical strip


 * $1 < \Re(s) \leq 1$

Then


 * $s \in \left\{{-2,-4,-6,\dots}\right\}$

These are called the trivial zeros of $\zeta$.

Proof
First we note that by Zeroes of the Gamma Function, $\Gamma$ has no zeros on $\C$.

Therefore, the Definition:Completed Riemann Zeta Function


 * $\displaystyle \xi(s) = \frac 12 s(s-1) \pi^{-s/2} \Gamma\left(\frac s2\right) \zeta(s)$

has the same zeros as $\zeta$.

Additionally by Functional Equation for Riemann Zeta Function, we have $\xi(s) = \xi(1-s)$ for all $s \in \C$.

Therefore if $\zeta(s) \neq 0$ for all $s$ with $\Re(s) > 1$ then also $\zeta(s) \neq 0$ for all $s$ with $\Re(s) < 0$.

Let us consider $\Re(s) > 1$. We have:


 * $\displaystyle \zeta(s) = \prod_{p} \frac 1 {1-p^{-s}}$

where here and in the following $p$ ranges over the primes.

Therefore, we have:
 * $\displaystyle \zeta(s) \prod_{p} \left({ 1-p^{-s} }\right) = 1$

All of the factors of this infinite product can be found in the product:


 * $\displaystyle \prod_{n=2}^\infty \left({ 1-n^{-s} }\right)$

which converges absolutely since the zeta sum $\displaystyle \sum_{k=1}^\infty k^{-s}$ converges absolutely.

Hence:


 * $\displaystyle \prod_{p} \left({ 1-p^{-s} }\right)$

converges absolutely, and so by the fact that:


 * $\displaystyle \zeta(s) \prod_{p} \left({ 1-p^{-s} }\right) = 1$

we know $\zeta(s)$ can't possibly be zero for any point in the region in question.