Bernoulli Number in terms of Euler Numbers
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Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
\(\ds B_{2 n}\) | \(=\) | \(\ds \frac {2n} {2^{2 n} \left({2^{2 n} - 1}\right)} \left({\sum_{k \mathop = 1}^n \dbinom {2 n - 1} {2 k - 1} E_{2 n - 2 k} }\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2n} {2^{2 n} \left({2^{2 n} - 1}\right)} \left({\binom {2 n - 1} 1 E_{2 n - 2} + \binom {2 n - 1} 3 E_{2 n - 4} + \binom {2 n - 1} 5 E_{2 n - 6} + \cdots + 1}\right)\) |
where:
- $B_n$ denotes the $n$th Bernoulli number
- $E_n$ denotes the $n$th Euler number.
Proof
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 21$: Relationships of Bernoulli and Euler Numbers: $21.7$