Sum of Reciprocals of Cubes of Odd Integers Alternating in Sign

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Theorem

\(\displaystyle \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n} {\paren {2 n + 1}^3}\) \(=\) \(\displaystyle \frac 1 {1^3} - \frac 1 {3^3} + \frac 1 {5^3} - \frac 1 {7^3} + \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^3} {32}\)


Proof

By Half-Range Fourier Sine Series for $x \paren {\pi - x}$ over $\openint 0 \pi$:

$\displaystyle x \paren {\pi - x} = \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \paren {2 n + 1} x} {\paren {2 n + 1}^3}$

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


Setting $x = \dfrac \pi 2$:

\(\displaystyle \frac \pi 2 \paren {\pi - \frac \pi 2}\) \(=\) \(\displaystyle \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\sin \paren {2 n + 1} \frac \pi 2} {\paren {2 n + 1}^3}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\sin \paren {n + \frac 1 2} \pi} {\paren {2 n + 1}^3}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^3}\) Sine of Half-Integer Multiple of Pi
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} 4 \times \frac \pi 8\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^3}\)

whence the result.

$\blacksquare$


Sources