Bernoulli numbers: comparison to factorials

The asymptotic of the Bernoulli numbers and of the central binomial coefficients is well known : $$|B_{2n}|\sim 4\sqrt{\pi\,n}\,\left(\frac n{\pi\,e}\right)^{2n},\qquad\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi\,n}}$$

This implies that \begin{align} \frac{2|B_{2n}|}{n!\;\binom{2n}{n}}&\sim \frac{8\sqrt{\pi\,n}}{n!}\,\left(\frac n{\pi\,e}\right)^{2n}\frac{\sqrt{\pi\,n}}{2^{2n}}\\ &\sim \frac{8\;{\pi\,n}}{\sqrt{2\,\pi\, n}}\,\left(\frac en\right)^{n}\,\left(\frac {n^2}{4\,\pi^2\,e^2}\right)^n\\ &\sim 4\sqrt{2\,\pi\, n}\,\left(\frac {n}{4\,\pi^2\,e}\right)^n\\ \end{align} This asymptotic goes clearly to infinity and will become larger than $1$ for $n$ a little smaller than $4\,\pi^2\,e\approx 107$, more exactly for $n=103$ as indicated by Old John.

According to pari/gp on my laptop, we have:

$$\frac{2B_{206}.103!}{206!} = 1.488\dots,$$

or, in pari/gp notation:

$$2*bernreal(2*103)*factorial(103)/factorial(2*103) = 1.488\dots,$$

which seems to indicate that your proposed result fails at $n=103$, and probably for all $n>103$.

To expand my comment: as Euler proved, $\zeta(2n)=|(2\pi)^{2n}B_{2n}/(2\times (2n)!)|$; since $\zeta(2k)>1$, we get $|2B_{2n}/(2n)!|>4/(2\pi)^{2n}$. But $1/n!$ goes to $0$ much faster.