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Re: [PrimeNumbers] Convergent?

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  • Phil Carmody
    ... 3^ln(n) = 3^(log_3(n)*ln(3)) = n^ln(3) sum(1/n^ln(3)) converges as n^ln(3) dominates n*ln(n)^2 whose sum of reciprocals converges. = sum(1/p^ln(3)) also
    Message 1 of 3 , Jul 21, 2007
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      --- "Werner D. Sand" <Theo.3.1415@...> wrote:
      > The divergence of sum(1/2^ln n) and sum(1/2^ln p) can easily be proven.
      > How can be proven that sum(1/3^ln n) and sum(1/3^ln p) converge?
      > (IF they do. What are the limits?)

      3^ln(n) = 3^(log_3(n)*ln(3)) = n^ln(3)
      sum(1/n^ln(3)) converges as n^ln(3) dominates n*ln(n)^2 whose sum of
      reciprocals converges.

      => sum(1/p^ln(3)) also converges.

      The numerical value? Sum the start and integrate functions bounding the tail
      from both sides.

      Phil

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    • Werner D. Sand
      ... proven. ... the tail ... Thank you. Remains adding that the convergence radius r of sum(1/a^ln n) and sum(1/a^ln p) is r=a e corresponding to the unity in
      Message 2 of 3 , Jul 22, 2007
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        --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...>
        wrote:
        >
        > --- "Werner D. Sand" <Theo.3.1415@...> wrote:
        > > The divergence of sum(1/2^ln n) and sum(1/2^ln p) can easily be
        proven.
        > > How can be proven that sum(1/3^ln n) and sum(1/3^ln p) converge?
        > > (IF they do. What are the limits?)
        >
        > 3^ln(n) = 3^(log_3(n)*ln(3)) = n^ln(3)
        > sum(1/n^ln(3)) converges as n^ln(3) dominates n*ln(n)^2 whose sum of
        > reciprocals converges.
        >
        > => sum(1/p^ln(3)) also converges.
        >
        > The numerical value? Sum the start and integrate functions bounding
        the tail
        > from both sides.
        >
        > Phil
        >


        Thank you. Remains adding that the convergence radius r of sum(1/a^ln
        n) and sum(1/a^ln p) is r=a>e corresponding to the unity in Riemann's
        Zeta function.

        Werner

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        > Be a better Globetrotter. Get better travel answers from someone
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