Sum of Reciprocals of Squares of Odd Integers/Proof 6

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Theorem

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\) \(=\) \(\ds 1 + \dfrac 1 {3^2} + \dfrac 1 {5^2} + \dfrac 1 {7^2} + \dfrac 1 {9^2} + \cdots\)
\(\ds \) \(=\) \(\ds \dfrac {\pi^2} 8\)


Proof

Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:

$\map f x = \begin{cases} -\pi & : 0 < x \le \pi \\ x - \pi & : \pi < x < 2 \pi \end{cases}$

By Fourier Series: $-\pi$ over $\openint 0 \pi$, $x - \pi$ over $\openint \pi {2 \pi}$, its Fourier series can be expressed as:

$\ds \map f x \sim \map S x = -\frac \pi 4 + \frac 2 \pi \sum_{r \mathop = 0}^\infty \frac {\cos \paren {2 r + 1} x} {\paren {2 r + 1}^2} - \sum_{n \mathop = 1}^\infty \frac {2 - \paren {-1}^n \sin n x} n$


Consider the point $x = \pi$.

By Fourier's Theorem:

\(\ds \map S \pi\) \(=\) \(\ds \frac 1 2 \paren {\lim_{x \mathop \to \pi^+} \map f x + \lim_{x \mathop \to \pi^-} \map f x}\)
\(\ds \) \(=\) \(\ds \frac {-\pi + \paren {\pi - \pi} } 2\)
\(\ds \) \(=\) \(\ds -\frac \pi 2\)


Thus:

\(\ds -\frac \pi 2\) \(=\) \(\ds -\frac \pi 4 + \frac 2 \pi \sum_{r \mathop = 0}^\infty \frac {\cos \paren {2 r + 1} \pi} {\paren {2 r + 1}^2} - \sum_{n \mathop = 1}^\infty \frac {2 - \paren {-1}^n \sin n \pi} n\)
\(\ds \leadsto \ \ \) \(\ds -\frac \pi 4\) \(=\) \(\ds \frac 2 \pi \sum_{r \mathop = 0}^\infty \frac {\cos \paren {2 r + 1} \pi} {\paren {2 r + 1}^2}\) Sine of Multiple of Pi and simplifying
\(\ds \) \(=\) \(\ds \frac 2 \pi \sum_{r \mathop = 0}^\infty \frac {-1} {\paren {2 r + 1}^2}\) Cosine of Multiple of Pi and simplifying
\(\ds \leadsto \ \ \) \(\ds \frac {\pi^2} 8\) \(=\) \(\ds \sum_{r \mathop = 1}^\infty \frac 1 {\paren {2 r - 1}^2}\) multiplying both sides by $-\dfrac \pi 2$ and adjusting indices

$\blacksquare$


Sources

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