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Does sum log(p)/(p(p-1)) converge?

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  • Décio Luiz Gazzoni Filho
    According to PARI/GP, sum_{p prime} log(p)/(p*(p-1)) appears to converge. ? maxp = 100000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s %1 =
    Message 1 of 3 , Feb 9, 2005
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      According to PARI/GP,

      sum_{p prime} log(p)/(p*(p-1))

      appears to converge.

      ? maxp = 100000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s
      %1 = 0.7553566278090918349304694602
      ? maxp = 1000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s
      %2 = 0.7553656108009827466628435907
      ? maxp = 10000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s
      %3 = 0.7553665108289091347321424396
      ? maxp = 100000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s
      %4 = 0.7553666008318318260357569144

      Any thoughts?

      I might share my interest on this sum with the list, if anyone cares. Just a
      hint: 1/p^2 + 1/p^3 + 1/p^4 + ... = 1/(p(p-1)).

      Décio


      [Non-text portions of this message have been removed]
    • Décio Luiz Gazzoni Filho
      Folks, don t waste your time on this one. David Broadhurst pointed me again to his method to compute sum_{p prime} log(p)/p, and one of the intermediate steps
      Message 2 of 3 , Feb 9, 2005
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        Folks, don't waste your time on this one. David Broadhurst pointed me again to
        his method to compute sum_{p prime} log(p)/p, and one of the intermediate
        steps in his derivation is

        sum_{p prime} log(p)/p = gamma + sum_{p prime} log(p)/(p*(p-1)).

        Thus the sum converges, and due to David's work it is known to 10,000 digits
        of accuracy.

        I'm sorry for wasting your time and for being so stupid as to forget David's
        wonderful method after just a week.

        Décio

        On Wednesday 09 February 2005 13:42, you wrote:
        > According to PARI/GP,
        >
        > sum_{p prime} log(p)/(p*(p-1))
        >
        > appears to converge.
        >
        > ? maxp = 100000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s
        > %1 = 0.7553566278090918349304694602
        > ? maxp = 1000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s
        > %2 = 0.7553656108009827466628435907
        > ? maxp = 10000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s
        > %3 = 0.7553665108289091347321424396
        > ? maxp = 100000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s
        > %4 = 0.7553666008318318260357569144
        >
        > Any thoughts?
        >
        > I might share my interest on this sum with the list, if anyone cares. Just
        > a hint: 1/p^2 + 1/p^3 + 1/p^4 + ... = 1/(p(p-1)).
        >
        > Décio


        [Non-text portions of this message have been removed]
      • Décio Luiz Gazzoni Filho
        ... This formula is wrong. Please replace it by B3 = gamma + sum_{p prime} log(p)/(p*(p-1)) where B3 = log x - sum_{p prime
        Message 3 of 3 , Feb 9, 2005
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          On Wednesday 09 February 2005 14:18, you wrote:
          > Folks, don't waste your time on this one. David Broadhurst pointed me again
          > to his method to compute sum_{p prime} log(p)/p, and one of the
          > intermediate steps in his derivation is
          >
          > sum_{p prime} log(p)/p = gamma + sum_{p prime} log(p)/(p*(p-1)).

          This formula is wrong. Please replace it by

          B3 = gamma + sum_{p prime} log(p)/(p*(p-1))

          where B3 = log x - sum_{p prime <= x} log(p)/p as x goes to infinity.

          Décio


          [Non-text portions of this message have been removed]
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