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Re: [PrimeNumbers] product convergence

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  • Andrey Kulsha
    ... 27/77, then 27/91, etc - also converges. But does the product converge to zero, or to a small but finite number? ... It converges to zero as well as the
    Message 1 of 8 , Apr 2, 1938
      > Clearly the terms converge to 1, and the product - 3/5, then 3/7, then
      27/77, then 27/91, etc - also converges. But does the product converge to
      zero, or to a small but finite number?
      >
      > I have been unable to answer him definitively. Can anyone else here do so?

      It converges to zero as well as the product of (p-1)/p.

      Best regards,

      Andrey
    • tsr21
      Thanks Andrey. Is there a simple proof of this? Tim
      Message 2 of 8 , Apr 2, 1938
        Thanks Andrey. Is there a simple proof of this?

        Tim

        --- In primenumbers@yahoogroups.com, Andrey Kulsha <Andrey_601@...> wrote:
        >
        > > Clearly the terms converge to 1, and the product - 3/5, then 3/7, then
        > 27/77, then 27/91, etc - also converges. But does the product converge to
        > zero, or to a small but finite number?
        > >
        > > I have been unable to answer him definitively. Can anyone else here do so?
        >
        > It converges to zero as well as the product of (p-1)/p.
        >
        > Best regards,
        >
        > Andrey
        >
      • djbroadhurst
        ... See Hardy and Wright Sections 22.7 and 22.8 for the Mertens product prod(p
        Message 3 of 8 , Apr 3, 1938
          --- In primenumbers@yahoogroups.com,
          "tsr21" <timro21@...> wrote:

          > Is there a simple proof of this?

          See Hardy and Wright Sections 22.7 and 22.8
          for the Mertens product

          prod(p<x, (p-1)/p) ~ exp(-Euler)/log(x)

          where the product runs over primes.

          As Andrey remarked, your product tends to zero faster:

          prod(2<p<x, (p-2)/p) = O(1/log(x)^2)

          David
        • djbroadhurst
          ... So let s work out the constant, say K, in prod(2
          Message 4 of 8 , Apr 3, 1938
            --- In primenumbers@yahoogroups.com,
            "djbroadhurst" <d.broadhurst@...> wrote:

            > prod(2<p<x, (p-2)/p) = O(1/log(x)^2)

            So let's work out the constant, say K, in

            prod(2<p<x, (p-2)/p) ~ K/log(x)^2

            We should use the twin-prime constant

            C2 = prod (2<p, p*(p-2)/(p-1)^2) = 0.6601618158...

            and then use the square of Mertens' formula,
            remembering that the latter includes p = 2.

            K = C2*(exp(-Euler)*2)^2 =
            0.832429065661945278030805943531465575045445318077417053240894...

            Sanity check:

            default(primelimit,10^8);
            \p5
            P=1.;x=10^8;forprime(p=3,x,P*=1-2/p);print(P*log(x)^2);

            0.83242

            Looks OK to me...

            David
          • Kermit Rose
            ... Hmm.... And I thought I had proven that the product converged to zero. David, what is wrong with my Proof below which I had already sent to Tim? Kermit
            Message 5 of 8 , Apr 4, 1938
              primenumbers@yahoogroups.com wrote:
              >
              >
              >
              > Messages in this topic (5)
              > ________________________________________________________________________
              > 1b. Re: product convergence
              > Posted by: "djbroadhurst" d.broadhurst@... djbroadhurst
              > Date: Tue May 4, 2010 1:17 pm ((PDT))
              >
              >
              >
              > --- In primenumbers@yahoogroups.com,
              > "djbroadhurst" <d.broadhurst@...> wrote:
              >
              >
              >> prod(2<p<x, (p-2)/p) = O(1/log(x)^2)
              >>
              >
              > So let's work out the constant, say K, in
              >
              > prod(2<p<x, (p-2)/p) ~ K/log(x)^2
              >
              > We should use the twin-prime constant
              >
              > C2 = prod (2<p, p*(p-2)/(p-1)^2) = 0.6601618158...
              >
              > and then use the square of Mertens' formula,
              > remembering that the latter includes p = 2.
              >
              > K = C2*(exp(-Euler)*2)^2 =
              > 0.832429065661945278030805943531465575045445318077417053240894...
              >
              > Sanity check:
              >
              > default(primelimit,10^8);
              > \p5
              > P=1.;x=10^8;forprime(p=3,x,P*=1-2/p);print(P*log(x)^2);
              >
              > 0.83242
              >
              > Looks OK to me...
              >
              > David
              >
              >
              >


              Hmm.... And I thought I had proven that the product converged to zero.

              David, what is wrong with my "Proof" below which I had already sent to Tim?

              Kermit


              > *****************************************************
              >
              >
              >
              > Hello Tim.
              >
              >
              >
              > http://en.wikipedia.org/wiki/Infinite_product
              >
              >
              >
              > The infinite product 3/5 5/7 9/11 ...
              >
              > converges to zero.
              >
              >
              >
              > would converge to a positive number between 0 and 1 only if
              >
              > the infinite product
              >
              > 5/3 7/5 11/9 ..... converged to a positive number > 1.
              >
              >
              >
              > 5/3 7/5 11/9 ..... = (1 + 2/3) (1 + 2/5) (1 + 2/9) (1 + 2/11) (1 +
              > 2/15) (1 + 2/17) ...
              >
              >
              > 1 + 2/3 + 2/5 + 2/9 + 2 / 11 + .... =< (1 + 2/3) (1 + 2/5) (1 + 2/9)
              > (1 + 2/11) (1 + 2/15) (1 + 2/17) ... =< exp( 2/3 + 2/5 + 2/9 + 2/11
              >
            • djbroadhurst
              ... Now let x tend to infinity and I think that you will see that K/log(x)^2 vanishes, not so :-? David
              Message 6 of 8 , Apr 4, 1938
                --- In primenumbers@yahoogroups.com,
                Kermit Rose <kermit@...> wrote:

                > I thought I had proven that the product converged to zero.

                It does. I wrote:

                > prod(2<p<x, (p-2)/p) ~ K/log(x)^2
                > K = 0.832429065661945278030805943531465575045445318077417053240894...

                Now let x tend to infinity and I think that you
                will see that K/log(x)^2 vanishes, not so :-?

                David
              • djbroadhurst
                ... It began OK and then fizzled out. Here is an elementary proof. Proposition: The product prod(p 2, (p-2)/p) over primes p 2 vanishes. Proof: It suffices to
                Message 7 of 8 , Apr 7, 1938
                  --- In primenumbers@yahoogroups.com,
                  Kermit Rose <kermit@...> wrote:

                  > what is wrong with my "Proof"

                  It began OK and then fizzled out.
                  Here is an elementary proof.

                  Proposition: The product prod(p>2, (p-2)/p) over primes p>2 vanishes.

                  Proof: It suffices to show that the contrary proposition is absurd.
                  Suppose that prod(p>2, (p-2)/p) did not vanish.
                  Then, by taking logs, we would conclude that
                  sum(p>2, log(p) - log(p-2)) is finite. But
                  log(p) - log(p-2) > 2/p. Thus
                  sum(p>2, 1/p) would also be finite.
                  Yet that is easily proven to be absurd:
                  http://primes.utm.edu/infinity.shtml#punchline
                  Hence the product prod(p>2, (p-2)/p) does indeed vanish.

                  Comment: By the same argument, it follows that
                  the product prod(p>x, (p-x)/p) over primes p > x
                  vanishes for all real x > 0.

                  David
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