Sum of Reciprocals of Sixth Powers of Odd Integers

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Theorem

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^6}\) \(=\) \(\ds 1 + \dfrac 1 {3^6} + \dfrac 1 {5^6} + \dfrac 1 {7^6} + \dfrac 1 {9^6} + \cdots\)
\(\ds \) \(=\) \(\ds \dfrac {\pi^6} {960}\)


Proof

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^6}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^6} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^6}\)
\(\ds \) \(=\) \(\ds \frac 1 {64} \sum_{n \mathop = 1}^\infty \frac 1 {n^6} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^6}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\pi^6} {945}\) \(=\) \(\ds \frac 1 {64} \times \dfrac {\pi^6} {945} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^6}\) Riemann Zeta Function of 6
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^6}\) \(=\) \(\ds \dfrac {\pi^6} {945} - \frac 1 {64} \times \dfrac {\pi^6} {945}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {64 - 1} \pi^6} {945 \times 64}\)
\(\ds \) \(=\) \(\ds \dfrac {\pi^6} {960}\)

$\blacksquare$


Also presented as

This result can also be seen presented as:

$\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^6} = \dfrac {\pi^6} {960}$


Sources

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