Loading ...
Sorry, an error occurred while loading the content.

Re: Convergent?

Expand Messages
  • 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 1 of 3 , Jul 22, 2007
    • 0 Attachment
      --- 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

      ______________________________________________________________________
      ______________
      > Be a better Globetrotter. Get better travel answers from someone
      who knows. Yahoo! Answers - Check it out.
      > http://answers.yahoo.com/dir/?link=list&sid=396545469
      >
    Your message has been successfully submitted and would be delivered to recipients shortly.