Sum of Reciprocals of Squares Alternating in Sign/Proof 4

Theorem

$\ds \dfrac {\pi^2} {12}$

$=$

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

$\ds $

$=$

$\ds \frac 1 {1^2} - \frac 1 {2^2} + \frac 1 {3^2} - \frac 1 {4^2} + \cdots$

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

$=$

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

$\ds $

$=$

$\ds \map \eta 2$

Definition of Dirichlet Eta Function, and $\paren {-1}^2 = 1$

$\ds $

$=$

$\ds \paren {1 - 2^{1 - 2} } \map \zeta 2$

$\ds $

$=$

$\ds \frac 1 2 \map \zeta 2$

$\ds $

$=$

$\ds \frac {\pi^2} {12}$

$\blacksquare$