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?

As others have pointed out, that's not quite the definition of the zeta function.  The zeta function is in fact the unique meromorphic function that's equal to that wherever that exists.  (To prove uniqueness, you can use Taylor series and the theorem that such a function is equal on any disc where it exists to the Taylor series at the center.)
Regarding the given value, I'll give Riemann's original proof:
$$\Gamma(x) = \int_0^\infty\frac{t^{x-1}}{e^t}dt = \int_0^\infty\frac{(nt)^{x-1}}{e^{nt}}ndt = n^x\int_0^\infty \frac{t^{x-1}}{e^{nt}}dt$$
Therefore
$$\Gamma(x)\zeta(x) = \sum_{n=1}^\infty\frac{\Gamma(x)}{n^x} = \sum_{n=1}^\infty\int_0^\infty \frac{t^{x-1}}{e^{nt}}dt = \int_0^\infty t^{x-1}\sum_{n=1}^\infty\frac{1}{(e^{t})^n}dt = \int_0^\infty\frac{t^{x-1}}{e^{t} - 1}dt$$
Now for the slightly tricky part: this still isn't defined for $\Re[x] < 1$.  To get around this, what he did was use the identity $2i\sin(\pi x) = e^{i\pi x}-e^{-i\pi x}$ to get
$$2i\sin(\pi x)\Gamma(x)\zeta(x) = (e^{i\pi x}-e^{-i\pi x})\int_0^\infty\frac{t^{x-1}}{e^{t} - 1}dt = \int_0^\infty\frac{(e^{-i\pi}t)^{x-1}}{e^{t} - 1}dt - \int_0^\infty\frac{(e^{i\pi}t)^{x-1}}{e^{t} - 1}dt$$
(Note I've left $e^{\pm i\pi}$ unsimplified, since you'll get different values for each when x isn't an integer.)  What Riemann did was he treated this as a contour integral, replacing the hairpin turn at the origin with a tight loop - which doesn't change the value since the integrand is analytic everywhere in the origin's neighborhood.  So what you ultimately have is
$$\zeta(x) = \lim_{s\to x}\frac{\int_C\frac{(-t)^{s-1}}{e^t - 1}dt}{2i\sin(\pi s)\Gamma(s)}$$
Where the contour treats -1 as $e^{i\pi}$ on the way in, $e^{-i\pi}$ on the way out, and goes around the origin in a tight loop in a clockwise (i.e., negative) direction.  Fortunately, in the case of integers positive or negative, there's no branch cut, so the way in and the way out cancel, leaving only a tight loop that can be computed with the residue theorem.  So with that in mind,
$$\int_C\frac{(-t)^{s-1}}{e^t - 1}dt = -\int_C (-t)^{s-2}\frac{t}{e^t - 1}dt$$
It turns out that the coefficients of the Maclaurin expansion of $\frac{x}{e^x-1}$ are just the Bernoulli numbers divided by the factorials, and every other Bernoulli number starting with the fourth is zero - meaning that the coefficient of every odd exponent of the Maclaurin series but $x^1$ is zero, so where s is a negative even integer, the residue is zero, so the integral is zero.
So there's still one hurdle - the gamma function has poles at nonpositive integers, and $\sin(\pi x)$ is equal to zero.  That's easy enough to work out, though, using the property $s\Gamma(s) = \Gamma(s+1)$ (which if you didn't know already, integration by parts will show you is implied by the definition above), and the fact that x is a negative even integer,
$$\lim_{s\to x}\frac{\frac{1}{\Gamma(s)}}{\sin(\pi s)} = \lim_{s\to x}\left(\frac{\prod_{n=0}^{-x}(s-x-n)}{(s-x)\Gamma(s-x+1)}\cdot\frac{(s-x)}{\sin(\pi s)}\right) = \frac{(-x)!}{\pi}$$
What this is isn't important, only that now you have zero in the numerator and a nonzero denominator, so you have zero.
