# Sum of Reciprocals of Even Powers of Integers Alternating in Sign/Corollary

## Corollary to Sum of Reciprocals of Even Powers of Integers Alternating in Sign

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

 $\ds B_{2 n}$ $=$ $\ds \paren {-1}^{n + 1} \dfrac {\paren {2 n}!} {\paren {2^{2 n - 1} - 1} \pi^{2 n} } \sum_{j \mathop = 1}^\infty \paren {-1}^{j + 1} \frac 1 {j^{2 n} }$ $\ds$ $=$ $\ds \paren {-1}^{n + 1} \dfrac {\paren {2 n}!} {\paren {2^{2 n - 1} - 1} \pi^{2 n} } \paren {1 - \dfrac 1 {2^{2 n} } + \dfrac 1 {3^{2 n} } - \dfrac 1 {4^{2 n} } + \cdots}$

## Proof

 $\ds \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n - 1} - 1} \pi^{2 n} } {\paren {2 n}!}$ $=$ $\ds \sum_{j \mathop = 1}^\infty \paren {-1}^{j + 1} \frac 1 {j^{2 n} }$ Sum of Reciprocals of Even Powers of Integers Alternating in Sign $\ds \leadsto \ \$ $\ds \paren {-1}^{n + 1} B_{2 n}$ $=$ $\ds \dfrac {\paren {2 n}!} {\paren {2^{2 n - 1} - 1} \pi^{2 n} } \sum_{j \mathop = 1}^\infty \paren {-1}^{j + 1} \frac 1 {j^{2 n} }$ multiplying both sides by $\dfrac {\paren {2 n}!} {\paren {2^{2 n - 1} - 1} \pi^{2 n} }$ $\ds \leadsto \ \$ $\ds \paren {-1}^{2 n + 2} B_{2 n}$ $=$ $\ds \paren {-1}^{n + 1} \dfrac {\paren {2 n}!} {\paren {2^{2 n - 1} - 1} \pi^{2 n} } \sum_{j \mathop = 1}^\infty \paren {-1}^{j + 1} \frac 1 {j^{2 n} }$ multiplying both sides by $\paren {-1}^{n + 1}$ $\ds \leadsto \ \$ $\ds B_{2 n}$ $=$ $\ds \paren {-1}^{n + 1} \dfrac {\paren {2 n}!} {\paren {2^{2 n - 1} - 1} \pi^{2 n} } \sum_{j \mathop = 1}^\infty \paren {-1}^{j + 1} \frac 1 {j^{2 n} }$ $\paren {-1}^{2 n + 2} = 1$ as $2 n + 2$ is even

$\blacksquare$