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

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 $x = \dfrac \pi 4$:

Now we need to show that:
 * $\displaystyle \frac 1 {1^3} + \frac 1 {3^3} - \frac 1 {5^3} - \frac 1 {7^3} + \cdots = \sum_{n \mathop = 0}^\infty \frac {\sin \frac {n \pi} 2 + \cos \frac {n \pi} 2} {\paren {2 n + 1}^3}$

For $n \equiv 0 \pmod 4$:
 * $\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \sin 2 k \pi + \cos 2 k \pi = 1$

For $n \equiv 1 \pmod 4$:
 * $\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \map \sin {2 k \pi + \dfrac \pi 2} + \map \cos {2 k \pi + \dfrac \pi 2} = 1$

For $n \equiv 2 \pmod 4$:
 * $\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \map \sin {2 k \pi + \pi} + \map \cos {2 k \pi + \pi} = -1$

For $n \equiv 3 \pmod 4$:
 * $\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \map \sin {2 k \pi + \dfrac {3 \pi} 2} + \map \cos {2 k \pi + \dfrac {3 \pi} 2} = -1$

This shows that the sum is alternating in sign in pairs.

Hence the result.