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Math Help - Sum of complex series

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    Sum of complex series

    Determine the sum of the series:

    \sum_{n=1}^{\infty}(-1)^n/(36n^2-1)
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  2. #2
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    Quote Originally Posted by moolimanj View Post
    Determine the sum of the series:

    \sum_{n=1}^{\infty}(-1)^n/(36n^2-1)
    A little late.

    Define the function,
    g(z) = \pi f(z) \csc \pi z = \frac{\pi \csc \pi z}{36z^2 - 1}

    Then,
    \sum_{n=-\infty}^{\infty}(-1)^n f(n) = -\sum_{k=1}^{m} \mbox{res}(g,a_k)

    Where a_1,a_2,...,a_m are the poles of f(z) = \frac{1}{36z^2 - 1} = \frac{1}{36\left(z - \frac{1}{6}\right)\left( z + \frac{1}{6} \right)}

    Thus, a_{1,2} = \pm \frac{1}{6}

    Then,
    \mbox{res}(g,\pm 1/6) = \lim_{z\to \pm 1/6} \left(z \mp \frac{1}{6} \right) \cdot \frac{\pi \csc \pi z}{36 \left( z - \frac{1}{6} \right) \left( z + \frac{1}{6} \right)}

    Thus,
    \sum_{k=1}^2 \mbox{res}(g,\pm 1/6) = - \frac{\pi}{12}.

    Which means,
    \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{36n^2 - 1} = \frac{\pi}{12}

    Thus,
    2\sum_{n=0}^{\infty} \frac{(-1)^n}{36n^2 - 1} = \frac{\pi}{12}

    Thus,
    \sum_{n=0}^{\infty} \frac{(-1)^n}{36n^2-1} = \frac{\pi}{24}
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