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21440product convergence

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  • Kermit Rose
    May 5, 2010
      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
      >
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