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:


 * $\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} = \frac {\pi^2} 8$