Why $\zeta(-2) $ is not $\sum_{n=1}^{\infty}\frac{1}{n^{-2}}$?

Question

Let $\zeta(s)= \sum_{n=1}^{\infty}\frac{1}{n^{s}}$ a standard formula.

I'm confused if you tell me: does this series: $\sum_{n=1}^{\infty}\frac{1} {n^{s}}$ converge?

I will answer you: this series is divergent. But if you say: $\zeta(-2)$ it will be: $\zeta(-2)= \sum_{n=1}^{\infty}\frac{1}{n^{-2}}=0$. Will be convergent. So why ?

Answer 1

See this, which gives a functional equation for $\zeta{(s)}$: $$\zeta{(s)} = 2^s \pi^{s-1} \sin{\left(\frac{\pi s}{2}\right)} \Gamma{(1-s)} \zeta{(1-s)}$$ All negative even integers are "trivial zeros" of the zeta function, because $\sin{\left(\frac{\pi s}{2}\right)}$ would equal $0$.

Answer 2

When people say "zeta function has zeros on negative even integers", they are talking about the analytic continuation of the Riemann zeta function, and the naive formula only works for Re(s)>1