Why are the trivial zeros negative even integers?

[beyond]

$$\varsigma(s) = \sum^{\infty}_{n=1}n^{-s}$$

If you substitute a trivial zero, let's say -2. Wouldn't it be

$$\varsigma(s) = \sum^{\infty}_{n=1} = 1^2 + 2^2 + 3^2 + 4^2 + . . .$$

How would this series be equals to zero?

Thanks

 

[antm88]

The formula that you provided is only valid for complex s with real parts greater than 1. Read the wiki page for more information:
http://en.wikipedia.org/wiki/Riemann_zeta_function
Anthony

 

[beyond]

So that means that the Riemann Hypothesis is based on the functional equation instead?

 

[rochfor1]

You should also look into analytic continuation.

 

[fourier jr]

& the bernoulli numbers, and the function $\xi$