Convergent Series

  • Thread starter Hertz
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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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  • #2
mathman
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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? $$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
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 $$\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}=\frac{\pi^2}{8}$$
 
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Thanks!
 
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WWGD
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I know the proof is written in detail in Dunham's "Journey Through Genius".
 

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