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Riemann zeta function

  1. Oct 11, 2008 #1
    [tex]\zeta (s)= \frac{1}{(1-2^{1-s})} \sum_{n=0}^{\infty} \frac {1}{(2^{n+1})} \sum_{k=0}^{n}(-1)^k{n \choose k}(k+1)^{-s} [/tex]

    Is the main problem with trying to prove the hypothesis algebraically boil down to the fact that s is an exponent to a "k" term? Would a derivation of the function that had no k terms to an exponent s, be at all helpful to an algebraic approach to the hypothesis?
  2. jcsd
  3. Oct 13, 2008 #2
    This problem gives me a headache!
    Last edited: Oct 14, 2008
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