Trivial Zeroes of Riemann Zeta Function are Even Negative Integers

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Theorem

Let $\rho = \sigma + i t$ be a zero of the Riemann zeta function not contained in the critical strip:

$0 \le \map \Re s \le 1$

Then:

$s \in \set {-2, -4, -6, \ldots}$

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


Proof


This needs considerable tedious hard slog to complete it.
In particular: First it needs to be established that those points are in fact zeroes. This follows directly and trivially from Riemann Zeta Function at Non-Positive Integers, but needs to be made explicit here.
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First we note that by Zeroes of Gamma Function, $\Gamma$ has no zeroes on $\C$.

Therefore, the completed Riemann zeta function:

$\map \xi s = \dfrac 1 2 s \paren {s - 1} \pi^{-s/2} \map \Gamma {\dfrac s 2} \, \map \zeta s$

has the same zeroes as $\zeta$.


Additionally by Functional Equation for Riemann Zeta Function, we have:

$\map \xi s = \map \xi {1 - s}$

for all $s \in \C$.

Therefore if:

$\map \zeta s \ne 0$

for all $s$ with $\map \Re s > 1$, then also:

$\map \zeta s \ne 0$

for all $s$ with $\map \Re s < 0$.


Let us consider $\map \Re s > 1$.

We have:

$\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$

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

Therefore, we have:

$\ds \map \zeta s \prod_p \paren {1 - p^{-s} } = 1$

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

$\ds \prod_{n \mathop = 2}^\infty \paren {1 - n^{-s} }$

which converges absolutely.

This follows because from Convergence of P-Series:

$\ds \sum_{k \mathop = 1}^\infty k^{-s}$ converges absolutely.


Hence:

$\ds \prod_p \paren {1 - p^{-s} }$ converges absolutely.

So by the fact that:

$\ds \map \zeta s \prod_p \paren {1 - p^{-s} } = 1$

we know $\map \zeta s$ cannot possibly be zero for any point in the region in question.

$\blacksquare$


Also see


Sources

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