In the functional equation for Riemann's zeta function, what happens at 2n, n natural
I am a bit confused on the functional equation for Riemann's zeta function,
for when s is a non-negative even integer. For example, when s = 2, then since
sin( ) = 0 , or for s=0, sin(0) = 0
it would seem that
(s) = 0
which apparently is not true, for two reasons:
(a) (s) is supposed to agree, for s> 1, with the series
from n = 1 to infinity,
but this latter sum will always be greater than zero for s>1 (to continue with the example of s=2, /6 0)
(b) non-negative even integers are not among the trivial zeros of the Riemann zeta function.
Another way to ask my question is why the functional equation =0 for negative even integers but not for non-negative even integers.