Prove $B_n\le n! $ for Bell numbers

Question

How using induction it can be shown that:

$$B_n\le n! \;\;\;\;\;\;\;\;\;\left( n\in \mathbb N \right)$$

Where $B_n$ is the nth Bell number.

The base case is true, since $$1=B_0\le 0!=1 \;\;\;\;\;\;\;\;\;\text{and}\;\;\;\;\;\;\;\;\; 1=B_1\le 1!=1$$

Assume the relation holds for $n=k$ and consider $n=k+1$:

This is where I cannot continue, I tried to use the relation $B_{k+1}\ge kB_{k-1}+B_k$, but that did not help me.

You can use the recurrence here. – Dietrich Burde Mar 19 '20 at 19:19 — Dietrich Burde, Mar 19 '20 at 19:19

Chris Grossack · Accepted Answer · 2020-03-19T19:27:49.223

1

Hint:

Consider instead the identity (listed on the Wikipedia page):

$$B_{n+1} = \sum_{k=0}^n \binom{n}{k} B_k$$

Then by (strong) induction, we know each $B_k \leq k!$, so:

$$ B_{n+1} = \sum_{k=0}^n \binom{n}{k} B_k \leq \sum_{k=0}^n \binom{n}{k} k! = \sum_{k=0}^n \frac{n!}{(n-k)!} $$

Can you take it from here?

I hope this helps ^_^

edited Mar 19 '20 at 19:27

answered Mar 19 '20 at 19:21

Chris Grossack

Ah, now I see what I've missed in my own attempts (+1)! – mrtaurho Mar 19 '20 at 19:25
2

$$\sum_{k=0}^{n}\frac{n!}{\left(n-k\right)!}\le\sum_{k=0}^{n}n!=\left(n+1\right)n!=\left(n+1\right)!$$ right? – Mar 19 '20 at 19:26
Looks good to me! ^_^ – Chris Grossack Mar 19 '20 at 19:29
@ very simple and fast (+1), thanks. – Mar 19 '20 at 19:29

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