Loading ...
Sorry, an error occurred while loading the content.

Re: 4 limits

Expand Messages
  • 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 1 of 9 , Jul 13, 2007
    • 0 Attachment
      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 2 of 9 , Jul 14, 2007
      • 0 Attachment
        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 3 of 9 , Jul 14, 2007
        • 0 Attachment
          > 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 4 of 9 , Jul 16, 2007
          • 0 Attachment
            --- 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 5 of 9 , Jul 28, 2007
            • 0 Attachment
              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]
              >
            Your message has been successfully submitted and would be delivered to recipients shortly.