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Convergent Series

  1. Sep 9, 2014 #1
    Hi, Mathematica is telling me the value of this series, but I can't figure out how to do it on paper. Can someone please explain? $$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
     
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  3. Sep 9, 2014 #2

    mathman

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    http://ptrow.com/articles/Infinite_Series_Sept_07.htm
    Above describes the derivation of
    $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

    Your equation is readily derived from this. Add all the $$\frac{1}{(2n)^2}$$ terms to your series and then subtract that series as [itex](\sum_{n=1}^{\infty}\frac{1}{n^2})/4[/itex]. You end up with $$\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}=\frac{\pi^2}{8}$$
     
  4. Sep 9, 2014 #3
    Thanks!
     
    Last edited: Sep 9, 2014
  5. Sep 9, 2014 #4

    WWGD

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    I know the proof is written in detail in Dunham's "Journey Through Genius".
     
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