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Re: product

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  • Werner D. Sand
    I suspect the product 2/3 * 7/5 * 11/13 * 19/17 * ... to converge to 1, oscillating around 1.
    Message 1 of 11 , Dec 4, 2005
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      I suspect the product 2/3 * 7/5 * 11/13 * 19/17 * ... to converge
      to 1, oscillating around 1.
    • Kermit Rose
      From: Werner D. Sand Date: 12/04/05 04:58:46 To: primenumbers@yahoogroups.com Subject: [PrimeNumbers] Re: product I suspect the product 2/3 * 7/5 * 11/13 *
      Message 2 of 11 , Dec 4, 2005
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        From: Werner D. Sand
        Date: 12/04/05 04:58:46
        To: primenumbers@yahoogroups.com
        Subject: [PrimeNumbers] Re: product

        I suspect the product 2/3 * 7/5 * 11/13 * 19/17 * ... to converge
        to 1, oscillating around 1.


        Kermit says:
        With primes, anything might be possible.

        [Non-text portions of this message have been removed]
      • Dario Alpern
        ... This product is not very difficult to compute if you know about the function gamma (the extension of factorials to the complex domain). You are trying to
        Message 3 of 11 , Dec 5, 2005
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          --- In primenumbers@yahoogroups.com, "Mark Underwood"
          <mark.underwood@s...> wrote:
          >
          > --- In primenumbers@yahoogroups.com, "Jacques Tramu"
          > <jacques.tramu@e...> wrote:
          > >
          > > I have computed the product and found
          > > n= 200 000 000 : 0.9048530178
          > > n= 1 000 000 0000 : 0.9048042986
          > > n= 1 500 000 000 : 0.9047987573
          > >
          >
          > I was just averaging some figures around the n = 40,000 area and it
          > worked out to be around .9105. Looking a Jacque's findings the
          > product seems to be ever so slowly shrinking.
          >
          > Contrast this the product 1/2 * 4/3 * 5/6 * 8/7 * ....
          >
          > and this appears to be converging to .599070...
          >
          > Mark
          >

          This product is not very difficult to compute if you know about the
          function gamma (the extension of factorials to the complex domain).

          You are trying to find:

          k=inf (4k+1)(4k+4) k=inf 2
          Prod ------------ = Prod 1 - ------------
          k=0 (4k+2)(4k+3) k=0 (4k+2)(4k+3)

          This is clearly convergent. But what is the limit?

          k=n (4k+1)(4k+4) k=n (k+1/4)(k+1)
          Prod ------------ = Prod -------------- =
          k=0 (4k+2)(4k+3) k=0 (k+2/4)(k+3/4)

          gamma(k+5/4) gamma(k+2) gamma(2/4) gamma(3/4)
          = ------------------------- ---------------------
          gamma(k+6/4) gamma(k+7/4) gamma(1/4) gamma(1)


          = A(k) x B

          where A(k) is the first fraction and B the second.

          We are interested in the value of A(k) as k->inf.

          Fortunately it turns out that the limit is 1. This can be seen by
          using Stirling approximation.

          Since gamma(1) = 1 and gamma(1/2) = sqrt(pi) we finally get:

          k=inf (4k+1)(4k+4) sqrt(pi) * gamma(3/4)
          Prod ------------ = ----------------------
          k=0 (4k+2)(4k+3) gamma(1/4)

          This is about 0.5990701173677961037199612

          Best regards,

          Dario Alejandro Alpern
          Buenos Aires - Argentina
          http://www.alpertron.com.ar/ENGLISH.HTM
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