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Euler's Zeta function equality and the Fall from Infinity

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  • Mark Underwood
    Hi all Checking out the Prime Pages, I was astonished to see a relation that Euler came up with, concerning his Zeta function: Z(n) = 1/1^n + 1/2^n + 1/3^n +
    Message 1 of 4 , Jun 4, 2003
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      Hi all

      Checking out the Prime Pages, I was astonished to see a relation that
      Euler came up with, concerning his Zeta function:

      Z(n) = 1/1^n + 1/2^n + 1/3^n + 1/4^n + 1/5^n + 1/6^n ....


      = 2^n/(2^n - 1) x 3^n/(3^n - 1) x 5^n/(5^n - 1) x
      7^n/(7^n - 1) x 11^n/(11^n - 1) x ...

      I find this relationship of the natural numbers to the primes to be
      incredible and mystifying.

      Recently someone asked about a Zeta like function diverging or
      converging, which got me wondering about something: If the Zeta
      function is divergent for n = 1 (becomes infinite), and if it
      converges to a finite number when n = 2 ( Z(2) incredibly equals
      (pi^2)/6 ), that must imply that there is a value for n between 1
      and 2 as n increases at which the series somehow falls from an
      infinite sum to a finite sum. What value of n would that be, anyone?
      It strains my brains that the sum could fall from infinite to finite
      with (I presume) an infinitely small increment of the n exponent.

      Mark
    • Satoshi Tomabechi
      ... The zeta function is convergent for all complex numbers except for n=1. Satoshi Tomabechi
      Message 2 of 4 , Jun 5, 2003
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        "Mark Underwood" <marku606@...> wrote:

        > Z(n) = 1/1^n + 1/2^n + 1/3^n + 1/4^n + 1/5^n + 1/6^n ....
        >
        > (pi^2)/6 ), that must imply that there is a value for n between 1
        > and 2 as n increases at which the series somehow falls from an
        > infinite sum to a finite sum. What value of n would that be, anyone?

        The zeta function is convergent for all complex numbers except for n=1.

        Satoshi Tomabechi
      • mikeoakes2@aol.com
        In a message dated 05/06/03 06:26:35 GMT Daylight Time, marku606@yahoo.ca ... Why so? tan(theta) falls from infinite to finite for an infinitely small
        Message 3 of 4 , Jun 5, 2003
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          In a message dated 05/06/03 06:26:35 GMT Daylight Time, marku606@...
          writes:


          > It strains my brains that the sum could fall from infinite to finite
          > with (I presume) an infinitely small increment of the n exponent.
          >

          Why so?
          tan(theta) falls from infinite to finite for an infinitely small increment of
          theta from pi/2...

          Mike Oakes


          [Non-text portions of this message have been removed]
        • Jon Perry
          It strains my brains that the sum could fall from infinite to finite ... Infinitely or finitely? Consider 1/e. As e- 0 the expression has a value, at e=0 it
          Message 4 of 4 , Jun 5, 2003
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            '> It strains my brains that the sum could fall from infinite to finite
            > with (I presume) an infinitely small increment of the n exponent.
            > '

            Infinitely or finitely?

            Consider 1/e. As e->0 the expression has a value, at e=0 it is undefined.
            This is basic continuity theory.

            The behaviour of zeta(s) around s=1 is analysed at;

            http://numbers.computation.free.fr/Constants/Miscellaneous/zeta.html

            Jon Perry
            perry@...
            http://www.users.globalnet.co.uk/~perry/maths/
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