Sum of Reciprocals of Fourth Powers of Odd Integers

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Theorem

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^4}\) \(=\) \(\ds 1 + \dfrac 1 {3^4} + \dfrac 1 {5^4} + \dfrac 1 {7^4} + \dfrac 1 {9^4} + \cdots\)
\(\ds \) \(=\) \(\ds \dfrac {\pi^4} {96}\)


Proof

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^4}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^4} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^4}\)
\(\ds \) \(=\) \(\ds \frac 1 {16} \sum_{n \mathop = 1}^\infty \frac 1 {n^4} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^4}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\pi^4} {90}\) \(=\) \(\ds \frac 1 {16} \times \dfrac {\pi^4} {90} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^4}\) Riemann Zeta Function of 4
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^4}\) \(=\) \(\ds \dfrac {\pi^4} {90} - \frac 1 {16} \times \dfrac {\pi^4} {90}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {16 - 1} \pi^4} {90 \times 16}\)
\(\ds \) \(=\) \(\ds \dfrac {\pi^4} {96}\)

$\blacksquare$


Also presented as

This result can also be seen presented as:

$\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^4} = \dfrac {\pi^4} {96}$


Sources

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