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Why are the trivial zeros negative even integers?

  1. Dec 30, 2009 #1
    [tex]\varsigma(s) = \sum^{\infty}_{n=1}n^{-s}[/tex]

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

    [tex]\varsigma(s) = \sum^{\infty}_{n=1} = 1^2 + 2^2 + 3^2 + 4^2 + . . .[/tex]

    How would this series be equals to zero?

    Thanks
     
  2. jcsd
  3. Dec 30, 2009 #2
  4. Dec 31, 2009 #3
    So that means that the Riemann Hypothesis is based on the functional equation instead?
     
  5. Jan 1, 2010 #4
  6. Jan 1, 2010 #5
    & the bernoulli numbers, and the function [itex]\xi[/itex]
     
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