Sum of Reciprocals of Cubes of Odd Integers Alternating in Sign

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Theorem

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


Proof

By Fourier Series for $x \left({\pi - x}\right)$ over $\left[{0 \,.\,.\, \pi}\right]$:

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

for $x \in \left[{0 \,.\,.\, \pi}\right]$.


Setting $x = \dfrac \pi 2$:

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

whence the result.

$\blacksquare$


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