Sum of Euler Numbers by Binomial Coefficients Vanishes/Corollary
Jump to navigation
Jump to search
Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
\(\ds E_{2 n}\) | \(=\) | \(\ds -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\binom {2 n} 0 E_0 + \binom {2 n} 2 E_2 + \binom {2 n} 4 E_4 + \cdots + \binom {2 n} {2 n - 2} E_{2 n - 2} }\) |
where $E_n$ denotes the $n$th Euler number.
Proof
From Sum of Euler Numbers by Binomial Coefficients Vanishes we have:
$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$
If:
- $\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$
then:
\(\ds \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} + E_{2 n}\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds E_{2 n}\) | \(=\) | \(\ds - \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}\) |
$\blacksquare$