20051207, 19:47  #1 
Nov 2004
2^{4} Posts 
Why zeta(2n)=0 ?
I know this wouldn't even deserve to be posted, but could someone explain me why zeta(2n)=0 ? Regards.

20051208, 02:34  #2  
Nov 2005
2^{4}×3 Posts 
Quote:
Start with one version of the functional equation for the zeta function: zeta(1s) == 2 (2pi)^(s) cos(s*pi/2)gamma(s)zeta(s) See Equation 10 at http://eww2lx.wri.wolfram.com/RiemannZetaFunction.html or many other sources for this equation. Substitute s = 2n+1 where n is a positive integer to get zeta(2n) == 2 (2pi)^(s) cos(n*pi + 1/2)(2n+1)!*zeta(2n+1) cos(n*pi + 1/2) == 0 and zeta(2n+1) is finite, so you get zeta(2n) == 0 Hope this helps, John 

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