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Re: [PrimeNumbers] Re: zeta limit

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  • Andrey Kulsha
    ... 1/2^k + 1/3^k + ... + 1/p^k + ... just equals to Sum[MoebiusMu[n]*Log[Zeta[n*k]]/n, {n, 1, +Infinity}] which converges rapidly; so it can be used to
    Message 1 of 5 , Apr 6, 2007
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      > I have wondered if there is an exponent
      > k such that
      >
      > (1/2)^k + (1/3)^k + (1/5)^k + (1/7)^k + ... (1/p)^k + ...
      >
      > has a limit of one.

      1/2^k + 1/3^k + ... + 1/p^k + ...

      just equals to

      Sum[MoebiusMu[n]*Log[Zeta[n*k]]/n, {n, 1, +Infinity}]

      which converges rapidly; so it can be used to estimate the waned value of k as
      1.3994333287263303...

      Best,

      Andrey


      [Non-text portions of this message have been removed]
    • Andrey Kulsha
      ... 1.399433328726330318202807214745644327904727429484383941274765822888062492487247800233390475384227... (if you need more digits) Best, Andrey [Non-text
      Message 2 of 5 , Apr 6, 2007
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        > 1.3994333287263303...

        1.399433328726330318202807214745644327904727429484383941274765822888062492487247800233390475384227...

        (if you need more digits)

        Best,

        Andrey

        [Non-text portions of this message have been removed]
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