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Math Help - Series : binomial coefficients

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Series : binomial coefficients

    Evaluate \sum_{n=1}^\infty {2n \choose n}^{-1}.
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    Spoiler:

    For n\geq 1, we have \frac{1}{{2n\choose n}}=\frac{n!n!}{(2n)!}=\frac{\Gamma(n+1)\Gamma(n+1  )}{\Gamma(2n+1)} =\frac{n^2\Gamma(n)}{2n\Gamma(2n)}=\frac{n}{2}B(n,  n) where B is Euler's Beta function, hence \frac{1}{{2n\choose n}}=\frac{n}{2}\int_0^1 x^{n-1}(1-x)^{n-1}dx.

    Since everything is positive, we may interchange sum and integral:

    \sum_{n=1}^\infty \frac{1}{{2n\choose n}}=\frac{1}{2}\int_0^1\left(\sum_{n=1}^\infty  n(x(1-x))^{n-1}\right)dx

    hence

    \sum_{n=1}^\infty \frac{1}{{2n\choose n}} = \frac{1}{2}\int_0^1\frac{dx}{(1-x+x^2)^2}.

    Now, we can apply usual methods; Maple gives the value \frac{1}{3}+\frac{2}{9\sqrt{3}}\pi.


    Last edited by Laurent; April 21st 2010 at 05:17 PM. Reason: Deleted a wrong contour integration proof (integrals on sides don't cancel)
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Great! My solution as well.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Great! My solution as well.
    I'd be interested to see if there is a method by inverting the series expansion for arcsine.
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