Solve Convergent Series with Mathematica: Pi Squared/8

[Hertz]

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? $$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$

 

[mathman]

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 $(\sum_{n=1}^{\infty}\frac{1}{n^2})/4$. You end up with $$\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}=\frac{\pi^2}{8}$$

 

[Hertz]

Thanks!

 

[WWGD]

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