# Prove the following series

• Jun 5th 2014, 12:16 PM
simosamkun
Prove the following series
Hi everyone,

I've this problem:

http://s1.postimg.org/6ov4blb1b/11111.png
• Jun 5th 2014, 08:45 PM
Prove It
Re: Prove the following series
It's just algebra, don't be scared of it...

\displaystyle \begin{align*} \pi ^2 &= \frac{\pi ^2}{3} - 4 \left( - \frac{1}{1^2} - \frac{1}{2^2} - \frac{1}{3^2} - \dots \right) \\ \pi ^2 &= \frac{ \pi ^2 }{3} + 4 \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \right) \\ \frac{2 \pi ^2}{3} &= 4 \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \right) \\ \frac{1}{4} \left( \frac{2\pi ^2}{3} \right) &= \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \\ \frac{ \pi ^2 }{6} &= \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \end{align*}
• Jun 6th 2014, 05:07 AM
Plato
Re: Prove the following series
You have above a model, now use 2) to derive a series for $\dfrac{\pi^2}{12}$

2) $\displaystyle 0= \frac{\pi ^2}{3} - 4 \left( \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \dots \right)$

If you add the two together you get $\dfrac{\pi^2}{8}$