Riemann Zeta Function at Even Integers/Corollary

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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

$\blacksquare$


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