Sum of Reciprocals of Cubes of Odd Integers Alternating in Sign

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

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

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.

$\blacksquare$