Quoted By:
maybe if you use the trigonomitrical representation... cause you have
\[
\zeta(s) = sum_{n} n^{-s} = sum_{n} n( cos(a) - i sin(b) ) = sum_{n} n ( cos(1/2) - i sin(b) ) = sum_{n} n cos(1/2) - i sum_{n} n sin(b)
\]
so you will have
\[
cos(1/2) sum_{n} n != 0
\]
I don't know what do you think (i just did that with basic knowledge maybe I'm wrong)
