Sum of Reciprocals of Squares Alternating in Sign/Proof 2

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Theorem

\(\ds \dfrac {\pi^2} {12}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \dfrac {\paren {-1}^{n + 1} } {n^2}\)
\(\ds \) \(=\) \(\ds \frac 1 {1^2} - \frac 1 {2^2} + \frac 1 {3^2} - \frac 1 {4^2} + \cdots\)


Proof

\(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n^2}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2} - \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^2}\) separating odd positive terms from even negative terms
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2} - \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\ds \) \(=\) \(\ds \frac {\pi^2} 8 - \frac 1 4 \times \frac {\pi^2} 6\) Sum of Reciprocals of Squares of Odd Integers, Basel Problem
\(\ds \) \(=\) \(\ds \frac {\pi^2 \paren {3 - 1} } {24}\)
\(\ds \) \(=\) \(\ds \frac {\pi^2} {12}\)

$\blacksquare$