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Re: 4 limits

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  • 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 1 of 9 , Jul 12, 2007
      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 2 of 9 , Jul 13, 2007
        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 3 of 9 , Jul 14, 2007
          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 4 of 9 , Jul 14, 2007
            > 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 5 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 6 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
                >
                >
                >
                > 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
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                > [Non-text portions of this message have been removed]
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