Bernoulli Number in terms of Euler Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Then:

\(\displaystyle B_{2 n}\) \(=\) \(\displaystyle \frac {2n} {2^{2 n} \left({2^{2 n} - 1}\right)} \left({\sum_{k \mathop = 1}^n \dbinom {2 n - 1} {2 k - 1} E_{2 n - 2 k} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2n} {2^{2 n} \left({2^{2 n} - 1}\right)} \left({\binom {2 n - 1} 1 E_{2 n - 2} + \binom {2 n - 1} 3 E_{2 n - 4} + \binom {2 n - 1} 5 E_{2 n - 6} + \cdots + 1}\right)\)

where:

$B_n$ denotes the $n$th Bernoulli number
$E_n$ denotes the $n$th Euler number.


Proof


Sources