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Math Help - zeta function series

  1. #1
    Super Member Random Variable's Avatar
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    zeta function series

    I want to show that  \displaystyle \sum^{\infty}_{n=1} \big( \zeta(4n)-1 \big) = \frac{7- 2 \coth \pi}{8}

     = \displaystyle \sum^{\infty}_{n=1} \sum^{\infty}_{m=2} \frac{1}{m^{4n}}

    Because I'm summing over only positive values, I think I can switch the order of summation.

     \displaystyle = \sum^{\infty}_{m=2} \sum^{\infty}_{n=1} \frac{1}{m^{4n}} = \sum^{\infty}_{m=2} \sum^{\infty}_{n=1} \Big(\frac{1}{m^{4}}\Big)^{n}

     \displaystyle = \sum^{\infty}_{m=2} \frac{\frac{1}{m^{4}}}{1- \frac{1}{m^{4}}} = \sum^{\infty}_{m=2} \frac{1}{m^{4}-1}

     = \displaystyle \sum_{m=2}^{\infty} \Big( \frac{1}{4(m-1)} - \frac{1}{4(m+1)} - \frac{1}{2(m^{2}+1)} \Big)

     = \displaystyle \frac{1}{4} \sum_{m=2}^{\infty} \Big( \frac{1}{(m-1)} - \frac{1}{(m+1)} \Big) - \frac{1}{2}  \sum^{\infty}_{m=2} \frac{1}{(m^{2}+1)}


    which I think is OK since both series converge

    The first series is telescoping and evaluates to \displaystyle \frac{3}{2} .

    The second series is where I run into problems.

    EDIT:  \displaystyle  \sum^{\infty}_{m=0} \frac{1}{(m^{2}+1)}  = \sum^{\infty}_{m=0} \frac{1}{(m^{2}+1)} - \frac{3}{2}

     \displaystyle = \frac{1}{2i} \sum^{\infty}_{m=0}  \Big(\frac{1}{m-i} - \frac{1}{m+i} \Big) - \frac{3}{2}


    At this point I probably need to use the digamma function, but I really don't know.
    Last edited by Random Variable; April 13th 2011 at 10:51 AM.
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  2. #2
    Super Member girdav's Avatar
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    You can apply Parseval to the function t\mapsto e^{-t}.
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by girdav View Post
    You can apply Parseval to the function t\mapsto e^{-t}.
    Parseval's identity?
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  4. #4
    Super Member girdav's Avatar
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    Yes.
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