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Math Help - Riemann Zeta Function

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Riemann Zeta Function

    Given  \displaystyle \text{Re}(s)>0 and  \displaystyle x>0 , show  \displaystyle \zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{  x^{n-k}}{(1+x)^{n+1}}\frac{(-1)^k}{(k+1)^s} .
    Last edited by chiph588@; August 18th 2010 at 01:58 PM. Reason: Typo as Vlasev pointed out below
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  2. #2
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    Chiph, I believe that the series should be:

    \displaystyle \frac{1}{1-2^{1-s}}\sum _{n=0}^{\infty } \sum _{k=0}^n \frac{(-1)^k<br />
   \binom{n}{k} x^{n-k}}{(k+1)^s (x+1)^{n+1}}}

    Here is my solution:

    Spoiler:


    Notice that if k \ge n+1, the summand is 0, so the upper limit n is irrelevant. Let it sum to infinity and it is the same sum. Then we can switch the order of the summation signs since now the inner sum does not depend on n.

    <br />
\displaystyle = \sum_{n=0}^{\infty}\sum_{k=0}^{n} \binom{n}{k}\frac{x^{n-k}}{(1+x)^{n+1}}\frac{(-1)^k}{(1+k)^s}
    \displaystyle = \sum_{k=0}^{\infty}\sum_{n=0}^{\infty} \binom{n}{k}\frac{x^{n-k}}{(1+x)^{n+1}}\frac{(-1)^k}{(1+k)^s}
    \displaystyle = \sum_{k=0}^{\infty} \frac{(-1)^k}{x^k(1+k)^s(1+x)} \sum_{n=0}^{\infty}  \binom{n}{k}\left(\frac{x}{1+x}\right)^n
    \displaystyle = \sum_{k=0}^{\infty} \frac{(-1)^k}{x^k(1+k)^s(1+x)} \frac{\left(\frac{x}{1+x}\right)^k}{\left(1-\frac{x}{1+x}\right)^{k+1}}
    \displaystyle = \sum_{k=0}^{\infty} \frac{(-1)^k(1+x)x^k}{x^k(1+k)^s(1+x)} \frac{\left(\frac{1}{1+x}\right)^k}{\left(\frac{1}  {1+x}\right)^{k}}
    \displaystyle = \sum_{k=0}^{\infty} \frac{(-1)^k}{(1+k)^s}
    \displaystyle = \sum_{k=0}^{\infty} \frac{(-1)^{n+1}}{n^s}

    Multiply the last line by the fraction and you get the analytic continuation of \zeta(s) for real part of s > 0.

    The sum was done by the formula
    <br />
\displaystyle \sum_{n}\binom{n}{k}y^n = \frac{y^k}{(1-y)^{k+1}}

    here we sum over all values of n. The summand is zero for n < k.

    Last edited by Vlasev; August 18th 2010 at 01:09 PM.
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  3. #3
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    Where did you find this identity?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Vlasev View Post
    Where did you find this identity?
    I found this a while back. It was in a book in my school's library so I can't be of much help. Perhaps it's online.
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  5. #5
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    There is one online but it's just a special case of this one. It's for x = 1. They do reference the paper where it was derived and I'm sure they did it for arbitrary x. Maybe that's what you saw. Thanks though.

    I think the series could be derived with a straight application of Euler summation: Euler summation - Wikipedia, the free encyclopedia
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