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?

Other answers have addressed in detail the specifics of the zeta function.  This is just to add a simple note which, I think, goes to the heart of the OP's question as to what he or she is missing, namely how can it make sense to talk about the behavior of a function defined by an infinite series at a value where the series diverges.
The simplest way to understand this is with the geometric series
$$f(x)=1+x+x^2+x^3+\cdots$$
This series clearly diverges if $|x|\gt1$.  Yet it makes sense to say $f(2)=-1$, because of the identity
$$1+x+x^2+x^3+\cdots={1\over1-x}$$
which holds for all $|x|\lt1$.  The right hand side is clearly meaningful for all $x\not=1$; it provides what's called an analytic continuation of the function $f$.
Something similar (but more complicated) occurs with the zeta function:  The series definition $\zeta(s)=\sum n^{-s}$ makes no sense when the real part of $s$ is less than $1$, but there are identities with expressions that do.  One of the things to look forward to, when you take a course in complex analysis, is learning why, for functions like the geometric series and the zeta function, you don't get conflicting identities that give different values for the analytic continuation, which is why it makes sense to talk about "the" zeta function $\zeta(s)$.
