Sum of Reciprocals of Squares of Odd Integers/Proof 1

Theorem

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}$

$=$

$\ds 1 + \dfrac 1 {3^2} + \dfrac 1 {5^2} + \dfrac 1 {7^2} + \dfrac 1 {9^2} + \cdots$

$\ds $

$=$

$\ds \dfrac {\pi^2} 8$

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2}$

$=$

$\ds \map \zeta 2$

$\ds $

$=$

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^2} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}$

$\ds $

$=$

$\ds \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}$

$\ds $

$=$

$\ds \frac 1 4 \map \zeta 2 + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}$

Hence:

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}$

$=$

$\ds \frac 3 4 \map \zeta 2$

$\ds $

$=$

$\ds \frac 3 4 \cdot \frac{\pi^2} 6$

$\ds $

$=$

$\ds \frac {\pi^2} 8$

$\blacksquare$