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Math Help - Partial Sums

  1. #1
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    Partial Sums

    Given that the summation from 1 to infinity of 1/n^2 = pi^2/6, evaluate
    a) the summation from 0 to infinity of 1/(2n+1)^2
    b) the summation from 0 to infinity of 1/(4n+1)^2
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  2. #2
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    \sum_{n=1}^{+\infty}\frac{1}{n^2}\;=\;\frac{\pi^2}  {6}

    \sum_{n=1}^{+\infty}\frac{1}{(2n)^2}\;=\;\sum_{n=1  }^{+\infty}\frac{1}{4n^2}

    \sum_{n=1}^{+\infty}\frac{1}{(2n)^2}\;=\;\frac{1}{  4}\sum_{n=1}^{+\infty}\frac{1}{n^2}\;=\;\frac{\pi^  2}{24}

    \sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}\;=\;\sum_{n  =1}^{+\infty}\frac{1}{n^2}-\sum_{n=1}^{+\infty}\frac{1}{(2n)^2}

    \sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}\;=\;\frac{\  pi^2}{6}-\frac{\pi^2}{24}=\;\frac{\pi^2}{8}
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    I got beaten but I'll post anyway,

    \sum\limits_{n=1}^{\infty }{\frac{1}{(2n+1)^{2}}}=\sum\limits_{n=1}^{\infty }{\frac{1}{(2n+1)^{2}}}-\frac{1}{4}\sum\limits_{n=1}^{\infty }{\frac{1}{n^{2}}}+\frac{1}{4}\sum\limits_{n=1}^{\  infty }{\frac{1}{n^{2}}}, hence your sum equals \left( \sum\limits_{n=1}^{\infty }{\frac{1}{(2n+1)^{2}}}+\frac{1}{4}\sum\limits_{n=  1}^{\infty }{\frac{1}{n^{2}}} \right)-\frac{1}{4}\sum\limits_{n=1}^{\infty }{\frac{1}{n^{2}}} and this is \sum\limits_{n=1}^{\infty }{\frac{1}{n^{2}}}-\frac{1}{4}\sum\limits_{n=1}^{\infty }{\frac{1}{n^{2}}}=\frac{3}{4}\sum\limits_{n=1}^{\  infty }{\frac{1}{n^{2}}}=\frac{\pi ^{2}}{8}.

    You can do the second one in the same fashion.
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