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Re: [PrimeNumbers] Re: 4 limits

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  • Peter Kosinar
    ... 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
    Message 1 of 9 , Jul 14 7:24 AM
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      > 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
    • Werner D. Sand
      ... 0.008? ... original ... required ... sum). ... is ... 1000, ... terms ... 0.5. ... You are quite right. I integrated only numerically. Everything clear.
      Message 2 of 9 , Jul 16 12:20 AM
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        --- 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 3 of 9 , Jul 28 5:56 PM
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          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)
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          > 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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