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Trivial zeros of the Riemann zeta function

  1. Feb 10, 2010 #1
    Clearly I am missing something obvious here, but how is it that negative even numbers are zeros of the Riemann zeta function?

    For example:

    [tex]\zeta (-2)=1+\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+...=1+4+9+..[/tex]

    Which is clearly not zero. What is it that I am doing wrong?
  2. jcsd
  3. Feb 10, 2010 #2
    You are using the definition of zeta(s) for Re(s) > 1 with a number that has a real part smaller than or equal to 1.
  4. Feb 10, 2010 #3


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    You need not the function you posted, but its analytic continuation.
  5. Feb 10, 2010 #4
    Thanks! It all becomes clear.
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