# Sum of Reciprocals of Cubes of Odd Integers Alternating in Sign

## 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

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

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.

$\blacksquare$