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?

There is a way to prove that $\zeta(-2K) = 0$ :


*

*By definition of the Bernouilli numbers $\frac{z}{e^z-1}= \sum_{k=0}^\infty \frac{B_k}{k!}z^k$ is analytic on $|z| < 2\pi$

*Note that 
$ \frac{z}{e^z-1}-\frac{z}{2} = \frac{z}{2}\frac{e^{z/2}+e^{-z/2}}{e^{z/2}-e^{-z/2}}$ is an even function, therefore $\frac{z}{e^z-1}-1-\frac{z}{2}=\sum_{k=2}^\infty \frac{B_k}{k!}z^k$ is an even function, and $B_{2k+1}=0$ for  $k\ge 1$.

*For $Re(s) > 0$, let $\Gamma(s) = \int_0^\infty x^{s-1} e^{-x}dx$. It converges absolutely so it is analytic. 
Integrating by parts $\Gamma(s+1) = s \Gamma(s)$ providing the analytic continuation to $Re(s) \le 0$ : $\Gamma(s) = \frac{\Gamma(s+k)}{\prod_{m=0}^{k-1} s+m}$. 
Thus $\Gamma(s)$ is analytic on $\mathbb{C} \setminus -\mathbb{N}$ with poles at the negative integers where $\Gamma(s) \sim \frac{(-1)^k}{k!}\frac{1}{s+k}$

*With the change of variable $x = ny$ you have $\Gamma(s) n^{-s} = \int_0^\infty x^{s-1} e^{-nx}dx$ so that for $Re(s) > 1$ where everything converges absolutely $$\Gamma(s) \zeta(s) = \sum_{n=1}^\infty \int_0^\infty x^{s-1} e^{-nx}dx= \int_0^\infty x^{s-1}\sum_{n=1}^\infty  e^{-nx}dx=\int_0^\infty x^{s-2}\frac{x}{e^x-1}dx$$

*Note that $\frac{1}{s+k-1} = \int_0^1 x^{s-2+k}dx = \int_0^\infty  x^{s-2} x^{k}1_{x < 1}dx$ so that
$$\Gamma(s) \zeta(s)- \sum_{k=0}^{K}\frac{B_k}{k!}\frac{1}{s+k-1} =\int_0^\infty x^{s-2}\left(\frac{x}{e^x-1}-\sum_{k=0}^K \frac{B_k}{k!}x^k1_{x < 1}\right)dx \tag{1}$$
Now as $x \to 0$  :$\frac{x}{e^x-1}-\sum_{k=0}^K \frac{B_k}{k!}x^k\sim \frac{B_{K+1}}{(K+1)!}x^{K+1}$ and hence $(1)$ converges and is bounded for $ Re(s) > -K$, i.e. as $s \to -k$ : 
$$\frac{(-1)^k}{k!}\frac{1}{s+k} \zeta(s) \sim\Gamma(s) \zeta(s)\sim \frac{B_{k+1}}{(k+1)!}\frac{1}{s+k} $$
whence
$$\boxed{\zeta(-k) = (-1)^k\frac{B_{k+1}}{k+1} \implies \zeta(-2k) = 0, k \in \mathbb{N}^*}$$
