Sum of Reciprocals of Squares of Odd Integers/Proof 7

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

By Half-Range Fourier Cosine Series for Identity Function over $\openint 0 \pi$:

$\ds x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$

for $x \in \openint 0 \pi$.

We have that:

$\map f \pi = \map f {\pi - 2 \pi} = \map f {-\pi} = \pi$

and so:

$\map f {\pi^-} = \map f {\pi^+}$

Hence we can set $x = \pi$:

\(\ds \pi\)

\(=\)

\(\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} \pi} {\paren {2 n - 1}^2}\)

\(\ds \)

\(=\)

\(\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {-1} {\paren {2 n - 1}^2}\)

Cosine of Multiple of Pi

\(\ds \leadsto \ \ \)

\(\ds \frac {\pi^2} 8\)

\(=\)

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

simplification

$\blacksquare$

Sources

• 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text I$: $6$.