On the Bell Numbers

Question

Edit (first version was incorrectly stated. Thank you Douglas and others for your corrections) Let $B_n$ be the $n$th Bell number (the number of partitions of a set with $n$ members). For each $n > 3$, I have a set $A_n$ of size $|A_n|=B_n$. I then have a subset $A'_n \subset A_n$ where $|A'_n|=B_n-B_{n-1}$. I would like to say something about the size of $A'_n$ relative to the size of $A_n$. For instance, it seems that $lim_{n \to \infty} \frac{B_{n-1}}{B_n}=0$. Can I make a stronger statement about the ratio of successive Bell numbers? How can I formalize the statement "for sufficiently large $n$, most of $A_n$ is in $A'_n$."

Answer 1

It's easy to see that $B_n \ge 2 B_{n-1}$ since we always have a choice of whether to add $n$ to the same part as $n-1$ or not. Since the number of parts in a typical set partition of size $n-1$ grows, the choices for adding $n$ to a new or existing part grow, so

$$\lim_{n\to\infty} B_{n-1}/B_n = 0.$$ There are asymptotics in the Wikipedia article on the Bell numbers, but it may not be obvious how to work with the Lambert $W$-function in that expression, or how to bound $B_{n-1}/B_n$. A faster proof that the limit is $0$ can be obtained from Dobiński's formula, that $B_n$ is the $n$th moment of a Poisson distribution with mean $1$:

For any $c \in \mathbb R$, the Poisson distribution has positive probability of being greater than $c$. So, for large enough $n$, the $n$th moment $B_n$ is at least $c^n$. By Jensen's inequality, the moments satisfy

$$B_n^{\frac{n+1}{n}} \le B_{n+1}$$

$$c \le \sqrt[n]{B_n} \le \frac {B_{n+1}}{B_n}$$

Answer 2

A precise estimate of the relative ratios of consecutive Bell numbers is proved by Asai, Nobuhiro; Kubo, Izumi; Kuo, Hui- Hsiung in Acta Appl. Math. 63 (2000), no. 1-3, 7987: $$1\leq \frac{B(n)B(n+2)}{B^2(n+1)}\leq \frac{n+2}{n+1}$$