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Re: limit = 6?

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  • Werner D. Sand
    That s right. One could also say: q ~ p, for p/q ~ 1. Then 1/(q-p*ln q) ~ 1/(p-p*ln p) and int(1/(x-x*ln x)) = -ln (ln x - 1) - -inf. Thanks. Can you
    Message 1 of 3 , Aug 9, 2007
      That's right. One could also say: q ~ p, for p/q ~ 1. Then 1/(q-p*ln
      q) ~ 1/(p-p*ln p) and int(1/(x-x*ln x)) = -ln (ln x - 1) -> -inf.
      Thanks. Can you nevertheless find out p when sum(1/(q-p*ln(q)) < 6
      for the first time?

      Werner



      --- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@...> wrote:
      >
      > Does it even converge?
      >
      > If gap size at p is about ln(p) then (with all equalities
      > being "approximately equal to") q=p+ln(p) so q-p*ln(q)=p+ln(p)-p*ln
      (p+ln
      > (p)) is on the order of p*ln(p) and the improper integral of 1/[x*ln
      > (x)] diverges.
      >
      > It should it diverge to -oo about as fast as ln(ln(p)).
      >
      > Adam
      >
      > >
      > > Let p < q be consecutive prime numbers.
      > > Does sum(1/(q-p*ln q), p=2..inf, converge exactly to 6?
      > >
      >
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