Sum of Reciprocals of Squares of Odd Integers/Proof 2

Proof
Let $n$ be a positive integer.

We have:

We also have:

So we can deduce:


 * $\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {\left({2 n + 1}\right)^2} = \frac {\pi^2} 8$