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Re: zeta limit

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  • Mark Underwood
    ... Speaking of converging to one, 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
    Message 1 of 5 , Apr 6, 2007
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      > >
      > > Limit pn/Sum[Zeta(i)*(pi-p(i-1)),{i,2,n}]=1
      >
      > > 1) Does someone can prove the firs limit?
      >
      > Yes, it's trivial and uninteresting.
      >


      Speaking of converging to one, 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. I just tried k = sqrt(2), and the first million
      primes yields a sum of about .97486

      For comparison, I just tried the same thing on the natural numbers
      starting at two.

      (1/2)^k + (1/3)^k + (1/4)^k + (1/5)^k + (1/6)^k + ...

      If k = sqrt(3), the first million terms sums to about .99378


      Just some reversion, no need for comment. A reflective Easter, everyone.

      Mark



      .



      > > 2) Does someone can prove that the second limit converges?
      >
      > Yes, it's trivial and uninteresting.
      >
      > Exactly the same type of relation holds true for any sequence that
      converges to
      > 1, not just {Zeta(i)}. There are an infinitude of these, your
      relation doesn't
      > connect Zeta to the primes in any way.
      >
      > Take the fine structure constant alpha ~1/137, and replace Zeta(i) by
      > alpha/CF_i(alpha) where CF_i is the i-th continued fraction convergent.
      >
      > Congratulations, your expression connects prime gaps to the fine
      structure
      > constant.
      >
      > Not!
      >
      > Phil
      >
      > Phil
      >
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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 2 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 3 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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