Sum of Reciprocals of Odd Powers of Odd Integers Alternating in Sign

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Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Then:

\(\displaystyle \sum_{j \mathop = 0}^\infty \frac {\left({-1}\right)^j} {\left({2 j + 1}\right)^{2 n + 1} }\) \(=\) \(\displaystyle \left({-1}\right)^{n + 1} \frac {\pi^{2 n + 1} E_{2 n} } {2^{2 n + 2} \left({2 n}\right)!}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {1^{2 n + 1} } - \frac 1 {3^{2 n + 1} } + \frac 1 {5^{2 n + 1} } - \frac 1 {7^{2 n + 1} } + \cdots\)

where $E_n$ is the $n$th Euler number.


Corollary

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

\(\displaystyle E_{2 n}\) \(=\) \(\displaystyle \paren {-1}^{n + 1} \dfrac {2^{2 n + 2} \paren {2 n}!} {\pi^{2 n + 1} } \sum_{j \mathop = 0}^\infty \frac {\paren {-1}^j} {\paren {2 j + 1}^{2 n + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {-1}^{n + 1} \dfrac {2^{2 n + 2} \paren {2 n}!} {\pi^{2 n + 1} } \paren {\frac 1 {1^{2 n + 1} } - \frac 1 {3^{2 n + 1} } + \frac 1 {5^{2 n + 1} } - \frac 1 {7^{2 n + 1} } + \cdots}\)


Proof


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