4

I am trying to understand the behaviour of the Bernoulli numbers with respect to factorials, specifically I'd like to know whether it is true that, for all $n \in N$ with $n \ge 2$ we have $$ \left|\frac{2B_{2n}}{(2n)!}\right| < \frac{1}{n!} $$

harlekin
  • 8,740
  • I may be off, but the formula for $\zeta(2n)$ and the fact that $\zeta(2n)>1$ seem to disagree with your inequality – user8268 Sep 25 '13 at 14:17
  • I found that the following eleven questions are closely-related or almost the same questions: (1) https://math.stackexchange.com/questions/783503/, (2) https://math.stackexchange.com/questions/580748/, (3) https://math.stackexchange.com/questions/1273516/, (4) https://math.stackexchange.com/questions/2568817/, (5) https://math.stackexchange.com/questions/2257544/, (6) https://math.stackexchange.com/questions/783503/, – qifeng618 Sep 22 '21 at 03:34
  • (7) https://math.stackexchange.com/questions/3447276/, (8) https://math.stackexchange.com/questions/504814/, (9) https://math.stackexchange.com/questions/1739872/, (10) https://math.stackexchange.com/questions/3451797/, (11) https://math.stackexchange.com/questions/2107114/. – qifeng618 Sep 22 '21 at 03:35

4 Answers4

4

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.

Raymond Manzoni
  • 43,021
  • 5
  • 86
  • 140
  • 2
    The most useful asymptotic formula remains of course $;\displaystyle |B_{2n}|\sim\frac{2;(2,n)!}{(2,\pi)^{2n}}\quad$ since $\zeta(n)\sim 1$ as $n\to +\infty$. – Raymond Manzoni Sep 25 '13 at 21:43
2

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 $n>103$.

Old John
  • 19,569
  • 3
  • 59
  • 113
2

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.

user8268
  • 21,348
0

The even-indexed Bernoulli numbers $B_{2k}$ satisfy the double inequality \begin{equation}\label{Bernoulli-ineq} \frac{2(2k)!}{(2\pi)^{2k}} \frac{1}{1-2^{\alpha -2k}} \le |B_{2k}| \le \frac{2(2k)!}{(2\pi)^{2k}}\frac{1}{1-2^{\beta -2k}}, \quad k\in\mathbb{N}, \end{equation} where $\alpha=0$ and \begin{equation*} \beta=2+\frac{\ln(1-6/\pi^2)}{\ln2}=0.6491\dotsc \end{equation*} are the best possible in the sense that they cannot be replaced respectively by any bigger and smaller constants.

References

  1. H. Alzer, Sharp bounds for the Bernoulli numbers, Arch. Math. (Basel) 74 (2000), no. 3, 207--211; available online at https://doi.org/10.1007/s000130050432.
  2. Feng Qi, A double inequality for the ratio of two non-zero neighbouring Bernoulli numbers, Journal of Computational and Applied Mathematics 351 (2019), 1--5; available online at https://doi.org/10.1016/j.cam.2018.10.049.
  3. Ye Shuang, Bai-Ni Guo, and Feng Qi, Logarithmic convexity and increasing property of the Bernoulli numbers and their ratios, Revista de la Real Academia de Ciencias Exactas Fisicas y Naturales Serie A Matematicas 115 (2021), no. 3, Paper No. 135, 12 pages; available online at https://doi.org/10.1007/s13398-021-01071-x.
qifeng618
  • 1,691