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 $A'\_n$."



On the Bell Numbers
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](http://en.wikipedia.org/wiki/Bell_number), 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 


$$B\_n^{\frac{n+1}{n}} \le B\_{n+1}$$


$$c \le \sqrt[n]{B\_n} \le \frac {B\_{n+1}}{B\_n}$$


