Sum of Reciprocals of Squares Alternating in Sign/Proof 3

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Theorem

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


Proof

Let $f \left({x}\right)$ be the real function defined on $\left({-\pi \,.\,.\, \pi}\right)$ as:

$f \left({x}\right) = \pi^2 - x^2$


By Fourier Series: $\pi^2 - x^2$ over $\left({-\pi \,.\,.\, \pi}\right)$:

$\displaystyle \pi^2 - x^2 \sim \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n - 1} \frac {\cos n x} {n^2}$

for $x \in \left({-\pi \,.\,.\, \pi}\right)$.


Setting $x = 0$:

\(\displaystyle \pi^2 - 0^2\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n - 1} \frac {\cos 0} {n^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n - 1} \frac 1 {n^2}\) Cosine of Zero is One
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} {12}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n - 1} \frac 1 {n^2}\) simplification
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n + 1} } {n^2}\) rearrangement

$\blacksquare$


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