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  • Werner D. Sand
    Who can help me calculating up to 10 exact decimal places sum (1/n^(1-1/n) - 1/n) sum (1/p^(1-1/p) - 1/p) sum (ln(n) / n^2) sum (ln(p) / p^2) n=positive
    Message 1 of 9 , Jul 4, 2007
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      Who can help me calculating up to 10 exact decimal places



      sum (1/n^(1-1/n) - 1/n)

      sum (1/p^(1-1/p) - 1/p)

      sum (ln(n) / n^2)

      sum (ln(p) / p^2)



      n=positive integer, p=prime, each from 1(2) to infinity?



      WDS

















      [Non-text portions of this message have been removed]
    • Werner D. Sand
      Thanks for evaluations received. What I mean is: Can sum(ln(n)/n^2 ~ 0.93 reach 1? Can sum(ln(p)/p^2 ~ 0.49 reach 0.5? WDS ... [Non-text portions of this
      Message 2 of 9 , Jul 7, 2007
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        Thanks for evaluations received. What I mean is:



        Can sum(ln(n)/n^2 ~ 0.93 reach 1?

        Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?




        WDS



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






        [Non-text portions of this message have been removed]
      • Peter Kosinar
        Hello, ... Consider the inequality sum(ln(k)/k^2, k=A+1..infinity)
        Message 3 of 9 , Jul 9, 2007
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          Hello,

          > Can sum(ln(n)/n^2 ~ 0.93 reach 1?

          Consider the inequality

          sum(ln(k)/k^2, k=A+1..infinity) < integral(ln(k)/k^2, k=A..infinity)

          The right-hand side can be evaluated as (ln(A)+1)/A. Taking A=2 yields
          sum(ln(k)/k^2, k=4..infinity) < (ln(3)+1)/3 < 0.700
          Clearly,
          sum(ln(k)/k^2, k=1..3) = ln(2)/4 + ln(3)/9 < 0.296

          Summing these two inequalities shows that your sum is bounded above by
          0.996. This estimate is quite rough, though; the actual value of the
          infinite sum seems to be around 0.9375...

          > Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?

          Unfortunately, I don't know a nice trick for this one :-)

          Peter

          --
          [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
        • Adam
          Using a similar trick, one can verify the sum over the primes is strictly less than 0.5. One overestimates the tail end of the sum: sum(log(t)/t^2,t=p to 00,
          Message 4 of 9 , Jul 12, 2007
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            Using a similar trick, one can verify the sum over the primes is
            strictly less than 0.5. One overestimates the tail end of the sum:
            sum(log(t)/t^2,t=p to 00, t prime) > sum(log(t)/t^2,t=p to 00, all t,
            prime or not) > int(log(t)/t^2,t=p-1 to 00, any t). If you use
            p=1009 then the initial segment sum is
            about .4921002678856370264032818 while the error term (integral) is
            about .007852900246658049529330132 and these add to (and form an
            upper bound of) .4999531681322950759326119. So the sum over all
            primes is strictly less than 0.5.

            Adam

            --- In primenumbers@yahoogroups.com, Peter Kosinar <goober@...> wrote:
            >
            > Hello,
            >
            > > Can sum(ln(n)/n^2 ~ 0.93 reach 1?
            >
            > Consider the inequality
            >
            > sum(ln(k)/k^2, k=A+1..infinity) < integral(ln(k)/k^2,
            k=A..infinity)
            >
            > The right-hand side can be evaluated as (ln(A)+1)/A. Taking A=2
            yields
            > sum(ln(k)/k^2, k=4..infinity) < (ln(3)+1)/3 < 0.700
            > Clearly,
            > sum(ln(k)/k^2, k=1..3) = ln(2)/4 + ln(3)/9 < 0.296
            >
            > Summing these two inequalities shows that your sum is bounded above
            by
            > 0.996. This estimate is quite rough, though; the actual value of
            the
            > infinite sum seems to be around 0.9375...
            >
            > > Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?
            >
            > Unfortunately, I don't know a nice trick for this one :-)
            >
            > Peter
            >
            > --
            > [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ]
            134813278
            >
          • Adam
            Whoops! Of course those inequalities are the other way around!! One overestimates the tail end of the sum: sum(log(t)/t^2,t=p to 00, t prime)
            Message 5 of 9 , Jul 13, 2007
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              Whoops!

              Of course those inequalities are the other way around!!

              One overestimates the tail end of the sum:
              sum(log(t)/t^2,t=p to 00, t prime) < sum(log(t)/t^2,t=p to 00, all t,
              prime or not) < int(log(t)/t^2,t=p-1 to 00, any t).... etc.

              Adam

              --- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@...> wrote:
              >
              > Using a similar trick, one can verify the sum over the primes is
              > strictly less than 0.5. One overestimates the tail end of the sum:
              > sum(log(t)/t^2,t=p to 00, t prime) > sum(log(t)/t^2,t=p to 00, all
              t,
              > prime or not) > int(log(t)/t^2,t=p-1 to 00, any t). If you use
              > p=1009 then the initial segment sum is
              > about .4921002678856370264032818 while the error term (integral) is
              > about .007852900246658049529330132 and these add to (and form an
              > upper bound of) .4999531681322950759326119. So the sum over all
              > primes is strictly less than 0.5.
              >
              > Adam
              >
              > --- In primenumbers@yahoogroups.com, Peter Kosinar <goober@> wrote:
              > >
              > > Hello,
              > >
              > > > Can sum(ln(n)/n^2 ~ 0.93 reach 1?
              > >
              > > Consider the inequality
              > >
              > > sum(ln(k)/k^2, k=A+1..infinity) < integral(ln(k)/k^2,
              > k=A..infinity)
              > >
              > > The right-hand side can be evaluated as (ln(A)+1)/A. Taking A=2
              > yields
              > > sum(ln(k)/k^2, k=4..infinity) < (ln(3)+1)/3 < 0.700
              > > Clearly,
              > > sum(ln(k)/k^2, k=1..3) = ln(2)/4 + ln(3)/9 < 0.296
              > >
              > > Summing these two inequalities shows that your sum is bounded
              above
              > by
              > > 0.996. This estimate is quite rough, though; the actual value of
              > the
              > > infinite sum seems to be around 0.9375...
              > >
              > > > Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?
              > >
              > > Unfortunately, I don't know a nice trick for this one :-)
              > >
              > > Peter
              > >
              > > --
              > > [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ]
              > 134813278
              > >
              >
            • Werner D. Sand
              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? Werner ... [Non-text portions of
              Message 6 of 9 , Jul 14, 2007
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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?

                Werner


                --- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@...> wrote:
                >
                > Whoops!
                >
                > Of course those inequalities are the other way around!!
                >
                > One overestimates the tail end of the sum:
                > sum(log(t)/t^2,t=p to 00, t prime) < sum(log(t)/t^2,t=p to 00, all t,
                > prime or not) < int(log(t)/t^2,t=p-1 to 00, any t).... etc.
                >
                > Adam
                >
                > --- In primenumbers@yahoogroups.com, "Adam" a_math_guy@ wrote:
                > >
                > > Using a similar trick, one can verify the sum over the primes is
                > > strictly less than 0.5. One overestimates the tail end of the sum:
                > > sum(log(t)/t^2,t=p to 00, t prime) > sum(log(t)/t^2,t=p to 00, all
                > t,
                > > prime or not) > int(log(t)/t^2,t=p-1 to 00, any t). If you use
                > > p=1009 then the initial segment sum is
                > > about .4921002678856370264032818 while the error term (integral) is
                > > about .007852900246658049529330132 and these add to (and form an
                > > upper bound of) .4999531681322950759326119. So the sum over all
                > > primes is strictly less than 0.5.
                > >
                > > Adam
                > >
                > > --- In primenumbers@yahoogroups.com, Peter Kosinar <goober@> wrote:
                > > >
                > > > Hello,
                > > >
                > > > > Can sum(ln(n)/n^2 ~ 0.93 reach 1?
                > > >
                > > > Consider the inequality
                > > >
                > > > sum(ln(k)/k^2, k=A+1..infinity) < integral(ln(k)/k^2,
                > > k=A..infinity)
                > > >
                > > > The right-hand side can be evaluated as (ln(A)+1)/A. Taking A=2
                > > yields
                > > > sum(ln(k)/k^2, k=4..infinity) < (ln(3)+1)/3 < 0.700
                > > > Clearly,
                > > > sum(ln(k)/k^2, k=1..3) = ln(2)/4 + ln(3)/9 < 0.296
                > > >
                > > > Summing these two inequalities shows that your sum is bounded
                > above
                > > by
                > > > 0.996. This estimate is quite rough, though; the actual value of
                > > the
                > > > infinite sum seems to be around 0.9375...
                > > >
                > > > > Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?
                > > >
                > > > Unfortunately, I don't know a nice trick for this one :-)
                > > >
                > > > Peter
                > > >
                > > > --
                > > > [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ]
                > > 134813278
                > > >
                > >
                >




                [Non-text portions of this message have been removed]
              • 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 7 of 9 , Jul 14, 2007
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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 8 of 9 , Jul 16, 2007
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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 9 of 9 , Jul 28, 2007
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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
                      >
                      >
                      >
                      > sum (1/n^(1-1/n) - 1/n)
                      >
                      > sum (1/p^(1-1/p) - 1/p)
                      >
                      > sum (ln(n) / n^2)
                      >
                      > sum (ln(p) / p^2)
                      >
                      >
                      >
                      > n=positive integer, p=prime, each from 1(2) to infinity?
                      >
                      >
                      >
                      > WDS
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      > [Non-text portions of this message have been removed]
                      >
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