# Thread: Prove the following series

1. ## Prove the following series

Hi everyone,

I've this problem:

2. ## 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*}

3. ## 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 $\dfrac{\pi^2}{8}$