Question: The Riemann zeta function $ζ(s)$ is defined for all complex numbers $s ≠ 1$ with a simple pole at $s = 1$. It has zeros at the **negative even integers**, i.e., at $s = −2, −4, −6, ...$. 

My question: How one can obtain these roots.

Answer: 
These roots arise from that the Riemann zeta function satisfies the following *functional equation* 

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

Thus, when s=-2n, we get that 

$$
\zeta(-2n)=2^{-2n} \pi^{-(2n+1)} sin(-n\pi ) \Gamma(1+2n)\zeta(1+2n)
$$

which gives us 

$$
\zeta(-2n) = 0
$$ 
as $sin(-n \pi)=0 $ and $\Gamma(1+2n) $ is finite for all n $\in \mathbb{Z}^+ $.
$$\zeta(-2)=\sum\_{k=1}^\infty \frac{1}{k^{-2}}$$ 
Consider the example: Zeta at negative integer $-2$. To me it is never zero, it is ever growing.