Sum of Reciprocals of Squares of Odd Integers

Theorem

 * $\displaystyle 1 + \dfrac 1 {3^2} + \dfrac 1 {5^2} + \dfrac 1 {7^2} + \dfrac 1 {9^2} + \cdots = \sum_{n \mathop = 1}^\infty \frac{1}{\left({2n-1}\right)^2} = \dfrac {\pi^2} 8$