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Infinite Zeta Function Sum

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the numerical value of [tex]\sum_{k=0}^{\infty} (\zeta(-k))[/tex]


    2. Relevant equations



    3. The attempt at a solution

    I have no idea how to get a numerical value for this sum.
     
  2. jcsd
  3. Jan 23, 2010 #2
    The sum [tex]\sum_k=0^\infty (\zeta(1-2k))[/tex] is equal to [tex]\sum_k=0^\infty (-B_{2k}\2k)[/tex]. I hope this helps. Thanks
     
  4. Jan 23, 2010 #3
    I forgot to mention that B represents the Bernoulli numbers.
     
  5. Jan 23, 2010 #4
    I think that

    [tex]\zeta(1 - 2k)[/tex] = [tex]\frac{(-1)^{2k-1} B_{2k}}{2k}[/tex].

    At least, that's what's in one of my books. Also, is that first sum correct? It looks like you're supposed to use the identity

    [tex]\frac{x}{e^x - 1} = \sum_{n=0}^\infty \frac{B_nx^n}{n!}[/tex]

    but you need another k in the sum's denominator.

    Petek
     
  6. Jan 23, 2010 #5
    Thank you for helping. Sorry about the first sum; I typed it in wrong.
     
  7. Jan 24, 2010 #6
    Because of the 1/k in the denominator, does this imply that [tex]\sum_{k=0}^{\infty} (\zeta(-k))[/tex] has no sum?
     
  8. Jan 24, 2010 #7
    I found that zeta(-k) is equal to B(n)/(((-1)^(n+1))*n), where B(n) is the Bernoulli numbers, implying that [tex]\sum_{k=0}^{\infty} (\zeta(-k))[/tex] is equal to ln(2)*[tex]\sum_{k=0}^{\infty} (B(n))[/tex]. Sorry about the formating.
     
  9. Jan 24, 2010 #8
    To make sure that we're solving the same problem, please post the sum that you're trying to evaluate (since one of your posts stated that the sum in your original post was inaccurate). Thanks!

    Petek
     
  10. Jan 24, 2010 #9
    The original sum I was trying to evaluate was [tex]\sum_{k=0}^{\infty} (\zeta(-k))[/tex].
     
  11. Feb 10, 2010 #10
    [tex]\sum_{n=1}^{\infty} 1/n^s[/tex]
     
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