Sum of Euler Numbers by Binomial Coefficients Vanishes/Corollary
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 | This article has been proposed for deletion. In particular: moved to parent page as suggested - thanks Please assess the validity of this proposal. (discuss) |
Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
| \(\displaystyle E_{2 n}\) | \(=\) | \(\displaystyle -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle -\paren {\binom {2 n} 0 E_0 + \binom {2 n} 2 E_2 + \binom {2 n} 4 E_4 + \cdots + \binom {2 n} {2 n - 2} E_{2 n - 2} }\) |
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:
| \(\displaystyle \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} + E_{2 n}\) | \(=\) | \(\displaystyle 0\) | |||||||||||
| \(\displaystyle E_{2 n}\) | \(=\) | \(\displaystyle - \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}\) |
$\blacksquare$
