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

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Theorem

\(\ds \frac 1 {1^3} + \frac 1 {3^3} - \frac 1 {5^3} - \frac 1 {7^3} + \cdots\) \(=\) \(\ds \frac {3 \pi^3 \sqrt 2} {128}\)


Proof

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

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

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


Setting $x = \dfrac \pi 4$:

\(\ds \frac \pi 4 \paren {\pi - \frac \pi 4}\) \(=\) \(\ds \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\sin \paren {2 n + 1} \frac \pi 4} {\paren {2 n + 1}^3}\)
\(\ds \) \(=\) \(\ds \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\map \sin {\frac {n \pi} 2 + \frac \pi 4} } {\paren {2 n + 1}^3}\)
\(\ds \) \(=\) \(\ds \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\sin \frac {n \pi} 2 \cos \frac \pi 4 + \cos \frac {n \pi} 2 \sin \frac \pi 4} {\paren {2 n + 1}^3}\) Sine of Sum
\(\ds \) \(=\) \(\ds \frac 8 \pi \times \frac 1 {\sqrt 2} \sum_{n \mathop = 0}^\infty \frac {\sin \frac {n \pi} 2 + \cos \frac {n \pi} 2} {\paren {2 n + 1}^3}\) Cosine of $45^\circ$, Sine of $45^\circ$
\(\ds \leadsto \ \ \) \(\ds \frac {3 \pi^2} {16} \times \frac \pi 8 \times \sqrt 2\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\sin \frac {n \pi} 2 + \cos \frac {n \pi} 2} {\paren {2 n + 1}^3}\)
\(\ds \) \(=\) \(\ds \frac {3 \pi^3 \sqrt 2} {128}\)


Now we need to show that:

$\ds \frac 1 {1^3} + \frac 1 {3^3} - \frac 1 {5^3} - \frac 1 {7^3} + \cdots = \sum_{n \mathop = 0}^\infty \frac {\sin \frac {n \pi} 2 + \cos \frac {n \pi} 2} {\paren {2 n + 1}^3}$

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 right hand side sum is alternating in sign in pairs.

Hence the result.

$\blacksquare$


Sources

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