Asymptotic Formula for Bernoulli Numbers
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Theorem
The Bernoulli numbers with even index can be approximated by the asymptotic formula:
- $B_{2 n} \sim \paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\dfrac n {\pi e} }^{2 n}$
where:
- $B_n$ denotes the $n$th Bernoulli number
- $\sim$ denotes asymptotically equal.
Proof
\(\ds \lim_{n \mathop \to \infty} \frac {B_{2 n} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} }\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {\paren {-1}^{n + 1} \paren {2 n}! \map \zeta {2 n} } {2^{2 n - 1} \pi^{2 n} \paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} }\) | Riemann Zeta Function at Even Integers | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {\paren {2 n}! \map \zeta {2 n} } {2^{2 n + 1} \sqrt {\pi n} \paren {\frac n e}^{2 n} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {2 \sqrt {\pi n} \paren {\frac {2 n} e}^{2 n} \map \zeta {2 n} } {2^{2 n + 1} \sqrt {\pi n} \paren {\frac n e}^{2 n} }\) | Stirling's Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map \zeta {2 n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
$\blacksquare$
Also rendered as
This result can also be seen expressed as:
- $B_{2 n} \sim \paren {-1}^{n + 1} 4 n^{2 n} \paren {\pi e}^{-2 n} \sqrt {\pi n}$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 21$: Asymptotic Formula for Bernoulli Numbers: $21.12$
- Weisstein, Eric W. "Bernoulli Number." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/BernoulliNumber.html