Trivial zeros of the Riemann Zeta function

A question that has been puzzling me for quite some time now:

Why is the value of the Riemann Zeta function equal to $0$ for every even negative number?

I assume that even negative refers to the real part of the number, while its imaginary part is $0$.

So consider $-2$ for example:

$f(-2) =
\sum_{n=1}^{\infty}\frac{1}{n^{-2}} =
\frac{1}{1^{-2}}+\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+\dots =
1^2+2^2+3^2+\dots =

\infty$

What am I missing here?

Answer

The Zeta function is defined as $\zeta(s)=\sum_{n\ge1}n^{-s}$ only for $s\in\mathbb{C}$ with $\Re(s)>1$!

The function on the whole complex plane (except a few poles) is the analytic continuation of that function.

On the Wikipedia page, you can find the formula:
$$\zeta(s)=\frac{2^{s-1}}{s-1}-2^s\int_0^\infty\frac{\sin(s\arctan t)}{(1+t^2)^{\frac{s}{2}}(e^{\pi t}+1)}dt$$

for $s\neq 1$. Maybe working on this integral for $s$ a negative integer will give you the result.