Sum of Reciprocals of Even Powers of Odd Integers

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Theorem

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.

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


Corollary

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.

\(\ds B_{2 n}\) \(=\) \(\ds \left({-1}\right)^{n + 1} \dfrac {2 \left({2 n}\right)!} {\left({2^{2 n} - 1}\right) \pi^{2 n} } \sum_{j \mathop = 1}^\infty \frac 1 {\left({2 j - 1}\right)^{2 n} }\)
\(\ds \) \(=\) \(\ds \left({-1}\right)^{n + 1} \dfrac {2 \left({2 n}\right)!} {\left({2^{2 n} - 1}\right) \pi^{2 n} } \left({1 + \dfrac 1 {3^{2 n} } + \dfrac 1 {5^{2 n} } + \dfrac 1 {7^{2 n} } + \cdots}\right)\)


Proof

\(\ds \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} }\) \(=\) \(\ds \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j}^{2 n} } + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }\)
\(\ds \) \(=\) \(\ds \frac 1 {2^{2 n} } \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} } + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }\)
\(\ds \leadsto \ \ \) \(\ds \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}\) \(=\) \(\ds \frac 1 {2^{2 n} } \times \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!} + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }\) Riemann Zeta Function at Even Integers
\(\ds \leadsto \ \ \) \(\ds \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }\) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!} - \frac 1 {2^{2 n} } \times \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}\)
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n} \pi^{2 n} } {2 \paren {2 n}!} - \paren {-1}^{n + 1} \dfrac {B_{2 n} \pi^{2 n} } {2 \paren {2 n}!}\)
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n} - 1} \pi^{2 n} } {2 \paren {2 n}!}\)

$\blacksquare$


Sources