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Math Help - Series ( tricky equation )

  1. #1
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    Series ( tricky equation )

    Having a lot of issue with proving this series:

    Given ∑ 1/n^2 converges to pi^2/6

    Show that ∑ 1/ (2n-1)^2 converges to pi^2/8

    Both equations have limits n=1 to infinity

    Any help would be greatly appreciated
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  2. #2
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    Re: Series ( tricky equation )

    Quote Originally Posted by youngpadawans View Post
    Having a lot of issue with proving this series:

    Given ∑ 1/n^2 converges to pi^2/6

    Show that ∑ 1/ (2n-1)^2 converges to pi^2/8

    Both equations have limits n=1 to infinity

    Any help would be greatly appreciated
    The sum you are trying to evaluate is the sum of reciprocals of squares of all the odd integers.

    You can evaluate this by using the sum of reciprocals of squares, and subtracting the sum of reciprocals of squares of the even integers.

    \displaystyle \begin{align*} \sum_{n = 1}^{\infty}{\frac{1}{(2n - 1)^2}} &= \sum_{n = 1}^{\infty}{\frac{1}{n^2}} - \sum_{n = 1}^{\infty}{\frac{1}{(2n)^2}} \\ &= \frac{\pi^2}{6} - \sum_{n = 1}^{\infty}{\frac{1}{4n^2}} \\ &= \frac{\pi^2}{6} - \frac{1}{4}\sum_{n = 1}^{\infty}{\frac{1}{n^2}} \\ &= \frac{\pi^2}{6} - \frac{1}{4}\cdot \frac{\pi^2}{6} \\ &= \frac{\pi^2}{6} - \frac{\pi^2}{24} \\ &= \frac{4\pi^2}{24} - \frac{\pi^2}{24} \\ &= \frac{3\pi^2}{24} \\ &= \frac{\pi^2}{8} \end{align*}
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  3. #3
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    Re: Series ( tricky equation )

    Thanks , this makes perfect sense!
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