Bell number, equivalent.

Question

The $n$-th Bell number $B_n$ can be defined by $\displaystyle e^{e^x-1}=\sum_{n=0}^{+\infty}\frac{B_n}{n!}x^n$ or $\displaystyle B_n=\frac 1e\sum_{k=0}^{+\infty}\frac{k^n}{k!}$ or $\displaystyle B_{n+1}=\sum_{k=0}^n\binom nkB_k$. My question : when $n\longrightarrow +\infty$, find a simple equivalent for $B_n$, and how to prove it. Thank you.

Answer 1

In the documentation of sequence $A000110 $ in $OEIS$, there a superb approximation proposed by Benoit Cloitre in $2002$. It write $$B_n \sim b^n e^{b-n-\frac{1}{2}} \sqrt{\frac{b}{b+n}}$$ where $$b\log(b)=n-\frac 12 \implies b=\frac{2 n-1}{2 W\left(\frac{1}{2} (2 n-1)\right)}$$ where $W(.)$ is Lambert function.

To give you an idea, I computed the logarithms of both for $n=10^k$ $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 0 & 0.00693073010 & 0\\ 1 & 11.6606419521 & 11.6611299296 \\ 2 & 266.357984292 & 266.357226410 \\ 3 & 4438.17691251 & 4438.17671459 \\ 4 & 63699.1791462 & 63699.1791119 \end{array} \right)$$