Explanation of trivial zeros of the Riemann Zeta Function

Question

Why do negative even numbers plugged into the Zeta function produce a zero? The Riemann Hypothesis implies that the non-trivial zeros are connected to the primes, so how does that fit with negative even numbers?

Answer 1

The functional equation of the zeta function, needed to extend anatically that function to $\;\Bbb C\setminus\{1\}\;$ and one of the most astonishingly beautiful equations in mathematics, is

$$\zeta(s)=2^s\pi^{s-1}\,\sin\frac{\pi s}2\,\Gamma(1-s)\,\zeta(1-s)$$

Well, now for $\;s=-2n\;,\;\;n\in\Bbb N\;$ , you get $\;\zeta(-2n)=0\;$ ...

Answer 2

For example, the zeta function at $-2$ is $1^2 + 2^2 + 3^2 + ...$ which is trivially equal to 0.

Similarly, the zeta function at $-4$ is $1^4 + 2^4 + 3^4 + ... $ which is also trivially equal to zero.

It is trivial to see that the same holds for the function at $-6$, $-8$, ... etc. For that reason, these are considered the trivial zeros of the Zeta function.