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Re: 4 limits

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  • Werner D. Sand
    ... 0.008? ... original ... required ... sum). ... is ... 1000, ... terms ... 0.5. ... You are quite right. I integrated only numerically. Everything clear.
    Message 1 of 9 , Jul 16, 2007
      --- In primenumbers@yahoogroups.com, Peter Kosinar <goober@...> wrote:
      >
      > > Fine trick, but what makes you sure the integral is not e.g. =
      0.008?
      > > Didn't you only shift the problem upon the integral?
      >
      > The integral can be evaluated explicitly, exactly as I did in my
      original
      > post. I just didn't find this method "nice trick"-y enugh, as it
      required
      > adding 168 terms first (as opposed to the two terms in the other
      sum).
      > It's easy to check that the integral int(log(t)/t^2, t=A..infinity)
      is
      > equal to (ln(A)+1)/A.
      >
      > Adam first added all the terms corresponding to primes smaller than
      1000,
      > which resulted in roughly 0.4921003 and then bounded the remaining
      terms
      > by the value of the integral corresponding to p=1009 (which means
      > A = p-1 = 1008). It's easy to see that (ln(1008)+1)/1008 < 0.007853.
      > These two values add up to 0.4999533, which is strictly less than
      0.5.
      >
      > Peter


      You are quite right. I integrated only numerically. Everything clear.
      Thanks to you both.

      Werner
      >
    • andrew_j_walker
      Hi, I m not sure if my email got through to you so I ll post here. I put a message on usenet a few years back
      Message 2 of 9 , Jul 28, 2007
        Hi, I'm not sure if my email got through to you so I'll post here.
        I put a message on usenet a few years back
        http://groups.google.com.au/group/sci.math.num-analysis/browse_thread/thread/d3f192b37f229eb3/ec74e93bc38819c3?lnk=st&q=walker+prime+sums&rnum=2#ec74e93bc38819c3

        with possible expansions for some prime sums, all unproven!
        You should also check Henri Cohen's paper (a dvi file)
        I linked to
        http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi
        which has some evaluations, especially the last one!

        Andrew

        --- In primenumbers@yahoogroups.com, "Werner D. Sand"
        <Theo.3.1415@...> wrote:
        >
        >
        > Who can help me calculating up to 10 exact decimal places
        >
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        >
        > sum (1/n^(1-1/n) - 1/n)
        >
        > sum (1/p^(1-1/p) - 1/p)
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        > sum (ln(n) / n^2)
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        > sum (ln(p) / p^2)
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        > n=positive integer, p=prime, each from 1(2) to infinity?
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        > WDS
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