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Re: Riemann Zeta Result

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  • djbroadhurst
    ... Surgeon General s warning: quitting Mathematica /now/ greatly reduces serious risks to your brain. I assume that x! means gamma(1+x). In which case, the
    Message 1 of 2 , Oct 22, 2011
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      --- In primenumbers@yahoogroups.com,
      Sebastian Martin Ruiz <s_m_ruiz@...> wrote:

      > Limit {x->0} (Zeta[2+(x!)^x]-Zeta[3])Log[x]/(x(Zeta[3]-Zeta[2+x^x]))=EulerGamma=0.57721...

      Surgeon General's warning: quitting Mathematica
      /now/ greatly reduces serious risks to your brain.

      I assume that x! means gamma(1+x). In which case, the value
      of the limit has /nothing/ to do with the Riemann zeta function.

      Try replacing "zeta" by "exp", or "log", any other function
      that is analytic in the neighbourhood of z=3, and you will
      get the /same/ result.

      Explanation: gamma(1+x) = 1 - Euler*x + O(x^2)

      David
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