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Thread: alternating series

  1. #1
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    infinite series

    Prove that

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

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


    and

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

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

    and finally what is $\displaystyle \sum_{k=0}^{\infty} \frac{1}{ \binom{4k}{2k} } $ ??
    Last edited by simplependulum; Aug 2nd 2009 at 01:17 AM.
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  2. #2
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    Quote Originally Posted by simplependulum View Post
    Prove that

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

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


    and

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

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

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

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

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


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