# Riemann Zeta Function at Even Integers/Corollary

## Corollary to Riemann Zeta Function at Even Integers

 $\displaystyle B_{2 n}$ $=$ $\displaystyle \left({-1}\right)^{n + 1} \dfrac {\left({2 n}\right)!} {2^{2 n - 1} \pi^{2 n} } \sum_{k \mathop = 1}^\infty \frac 1 {k^{2 n} }$ $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^{n + 1} \dfrac {\left({2 n}\right)!} {2^{2 n - 1} \pi^{2 n} } \left({1 + \frac 1 {2^{2 n} } + \frac 1 {3^{2 n} } + \cdots}\right)$

where:

$B_n$ are the Bernoulli numbers
$n$ is a positive integer.

## Proof

 $\displaystyle \left({-1}\right)^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\left({2 n}\right)!}$ $=$ $\displaystyle \zeta \left({2 n}\right)$ Riemann Zeta Function at Even Integers $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^\infty \frac 1 {k^{2 n} }$ Definition of Riemann Zeta Function $\displaystyle \leadsto \ \$ $\displaystyle \left({-1}\right)^{n + 1} B_{2 n}$ $=$ $\displaystyle \dfrac {\left({2 n}\right)!} {2^{2 n - 1} \pi^{2 n} } \sum_{k \mathop = 1}^\infty \frac 1 {k^{2 n} }$ multiplying both sides by $\dfrac {\left({2 n}\right)!} {2^{2 n - 1} \pi^{2 n} }$ $\displaystyle \leadsto \ \$ $\displaystyle \left({-1}\right)^{2 n + 2} B_{2 n}$ $=$ $\displaystyle \left({-1}\right)^{n + 1} \dfrac {\left({2 n}\right)!} {2^{2 n - 1} \pi^{2 n} } \sum_{k \mathop = 1}^\infty \frac 1 {k^{2 n} }$ multiplying both sides by $\left({-1}\right)^{n + 1}$ $\displaystyle \leadsto \ \$ $\displaystyle B_{2 n}$ $=$ $\displaystyle \left({-1}\right)^{n + 1} \dfrac {\left({2 n}\right)!} {2^{2 n - 1} \pi^{2 n} } \sum_{k \mathop = 1}^\infty \frac 1 {k^{2 n} }$ $\left({-1}\right)^{2 n + 2} = 1$ as $2 n + 2$ is even

$\blacksquare$