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Proving the golden key,

  1. Apr 24, 2006 #1
    Hello,
    I'm trying to prove that

    [tex]\sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p} (1-p^ {-s} )^ {-1} [/tex]

    I know why it is and a proof, but I'm actually looking for
    a different way to prove going backward and deriving the
    sum from the product of primes. Can you show me a way to do that?
    I'd like to start with...
    [tex] \prod_{p} (1-p^ {-s} )^ {-1} = ... [/tex]

    Thanks,

    p.s that -s above and then following -1 should be both exponents for the equation
    same with -s and -1 on the bottom, I'm not the master latex writer. Can somebody also tell me why it doesn't work?
     
    Last edited: Apr 24, 2006
  2. jcsd
  3. Apr 25, 2006 #2
    The golden key lives in the golden house with goldielocks. It is guarded by two dragons. Penitent man will pass.
     
  4. Apr 25, 2006 #3
    I will never pass... but! I wouldn't have to, if you show me the very proof :)
     
  5. Apr 25, 2006 #4

    shmoe

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    Homework Helper

    What proof do you know? The usual is to look at the finite product [tex]\prod_{p\leq x}(1-p^{-s})^{-1}[/tex] and expand using geometric series. Then compare with [tex]\sum_{n\leq x} n^{-s}[/tex], and show the difference the two goes to zero as x-> infinity.
     
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