# On the ratio $\frac{F_n}{B_n}$

One of the interesting limits that I came up with is:

$$\lim_{n\to\infty} \frac{F_{n}}{B_{n}}\;\;\;\;\;\;\;\;\;\; \left( n \in \mathbb N^+\right)$$

Where $$F_n$$ is the nth Fibonacci number and $$B_n$$ is the nth Bell number.

If $$n$$ is a natural odd number then it can be written as $$n=2k-1$$ , where $$k\in \mathbb N^+$$, Using Stirling's approximation for the double factorial denoted $$n!!=\left(2k-1\right)!!$$ and the relation $$B_{n}\ge n!!$$ we have: $$0<\frac{F_{n}}{B_{n}} <\frac{F_{n}}{n!!}\sim \frac{\left(\frac{1+\sqrt{5}}{2}\right)^{2k-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{2k-1}}{\sqrt{5}}\cdot\frac{2^{k}\sqrt{2\pi k}\left(\frac{k}{e}\right)^{k}}{\sqrt{4\pi k}\left(\frac{2k}{e}\right)^{2k}}<\frac{2\cdot2^{k}}{k^{k}}$$

Taking the limit as $$k \to \infty$$ and using squeeze theorem follows: $$\lim_{n\to\infty} \frac{F_{n}}{B_{n}}=0$$

Which means as $$n$$ gets larger,the fraction with the numerator counting the number of ways to tile a board of size $$1×n$$ with squares and dominoes of size $$1×1$$ and $$1×2$$ respectively and the dinominator counting all possible partitions of a set with cardinality $$n+1$$ gets smaller.

The same can be done for $$n$$ even. For more information refer to this link.

Note: I've already proved that for all $$k \in \mathbb N$$ the relation $$B_k\ge F_k$$ holds, using this we conclude that:$$0<\frac{F_{n}}{B_{n}}\le1$$

The question is that: does there exist a more elegant way to prove this convergence?

• $(3/2)^n < F_n \sim \frac{1}{\sqrt{5}} \left( \frac{1 + \sqrt{5}}{2} \right)^n < 2^n$ but I think you're already using something like that. Mar 19 '20 at 16:47
• Since both $F_n$ and $B_n$ are always positive, you don't need to give that lower bound for $F_n/B_n.$ Mar 19 '20 at 17:05
• @ kenta,then how should I use squeeze theorem?
– user715522
Mar 19 '20 at 17:11
• You already know $F_n/B_n>0$ Mar 19 '20 at 17:25
• @ Kenta S , you are totally right, I will edit that, thanks.
– user715522
Mar 19 '20 at 17:26

Sicne $$F_n\le 2^n$$ and $$B_n\ge n!!,$$ we can say that $$F_n/B_n\le 2^n/n!!.$$ Since it is easily proven inductively that for large enough $$n$$ ($$n\ge 21$$), $$n!!\ge 3^n,$$ we have that, for large enough $$n,$$ $$0 Now use Squeeze Theorem.
• I will prove the inequality when $n$ is even, as the odd case is exactly the same. First of all, $20!!\ge 3^20,$ as can be checked by a calculator. Now let $n>20$ be an even number, and assume the inequality for $n-2.$ We have that $n!!=n\cdot (n-2)!\ge n\cdot 3^{n-2}\ge 9\cdot 3^{n-2}=3^n.$ By induction, the inequality is proven for all even $n.$ Mar 19 '20 at 17:49
• Thanks, I just think it should be $n!!=n\cdot\left(n-2\right)!!$ instead of $n!!=n\cdot\left(n-2\right)!$