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

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