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Math Help - alternating series

  1. #1
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    infinite series

    Prove that

     \sum_{k=0}^{\infty} \frac{ (-1)^k}{ \binom{2k}{k} }

     = \frac{4}{5} [ 1 -  \frac{1}{\sqrt{5}} \ln( \frac{ 1 + \sqrt{5}}{2} ) ]


    and

      \sum_{k=0}^{\infty} \frac{1}{ \binom{2k}{k} }

     = \frac{ 2( \pi + 6\sqrt{3})}{9\sqrt{3}}

    and finally what is   \sum_{k=0}^{\infty} \frac{1}{ \binom{4k}{2k} }  ??
    Last edited by simplependulum; August 2nd 2009 at 02:17 AM.
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  2. #2
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    Quote Originally Posted by simplependulum View Post
    Prove that

     \sum_{k=0}^{\infty} \frac{ (-1)^k}{ \binom{2k}{k} }

     = \frac{4}{5} [ 1 - \frac{1}{\sqrt{5}} \ln( \frac{ 1 + \sqrt{5}}{2} ) ]


    and

     \sum_{k=0}^{\infty} \frac{1}{ \binom{2k}{k} }

     = \frac{ 2( \pi + 6\sqrt{3})}{9\sqrt{3}}

    and finally what is  \sum_{k=0}^{\infty} \frac{1}{ \binom{4k}{2k} } ??
    let f(\alpha)=1 + \sum_{k \geq 1} \frac{\alpha^k}{\binom{2k}{k}}, \ \ 0 \neq |\alpha| \leq 1. your question is to find f(-1), \ f(1), and \frac{f(1)+f(-1)}{2}. we have: \frac{1}{\binom{2k}{k}}=\frac{(k!)^2}{(2k)!}=\frac  {k \Gamma(k) \Gamma(k+1)}{\Gamma(2k+1)}=kB(k,k+1)=k \int_0^1 x^{k-1}(1-x)^k \ dx.

    hence: f(\alpha)=1 + \sum_{k \geq 1}k \alpha^k \int_0^1 x^{k-1}(1-x)^k \ dx= 1 + \int_0^1 \frac{1}{x} \sum_{k \geq 1} k(\alpha x - \alpha x^2)^k \ dx. therefore: f(\alpha)= 1 + \int_0^1 \frac{\alpha(1-x)}{(\alpha x^2 - \alpha x + 1)^2} \ dx=1 + \frac{1}{\alpha} \int_0^1 \frac{1-x}{(x^2 - x + \frac{1}{\alpha})^2} \ dx.

    finally substituting x \to x - \frac{1}{2} gives us: f(\alpha) = 1 + \frac{1}{\alpha} \int_0^{\frac{1}{2}} \frac{dx}{(x^2 + \beta)^2}, where \beta = \frac{1}{\alpha} - \frac{1}{4}.


    Example: let's find f(1) : we have f(1)=1 + \int_0^{\frac{1}{2}} \frac{dx}{(x^2 + \frac{3}{4})^2}. let x = \frac{\sqrt{3}}{2} \tan t. then \int_0^{\frac{1}{2}} \frac{dx}{(x^2 + \frac{3}{4})^2}=\frac{8 \sqrt{3}}{9} \int_0^{\frac{\pi}{6}} \cos^2 t \ dt=\frac{1}{3} + \frac{2 \sqrt{3} \pi}{27}. therefore: f(1)=\frac{4}{3} + \frac{2 \sqrt{3} \pi}{27}.
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