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

## 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.

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