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Thread: Riemann Zeta Function

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

    Given $\displaystyle \displaystyle \text{Re}(s)>0 $ and $\displaystyle \displaystyle x>0 $, show $\displaystyle \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@; Aug 18th 2010 at 12:58 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:

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

    Here is my solution:

    Spoiler:


    Notice that if $\displaystyle k \ge n+1$, the summand is 0, so the upper limit $\displaystyle n$ 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 $\displaystyle n$.

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

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

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

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

    Last edited by Vlasev; Aug 18th 2010 at 12:09 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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