# How we can know the bounds for the factorial $n!$

If we know that there exist a positive integer $$n$$ in an interval $$[a,b]$$. Here $$a,b$$ are real numbers. How we can know the bounds for the factorial $$n!$$.

• What's wrong with $b!$? – Arthur Dec 5 '19 at 13:29
• @Arthur: Here $a,b$ are real numbers. – Safwane Dec 5 '19 at 13:30
• And with $\lfloor a\rfloor!$ and $\lceil b\rceil!\,$? – Bernard Dec 5 '19 at 13:33
• @Gabe Yes you can: the other natural idea $a\cdot (a-1)\cdot \ldots$ is sometimes better (say $a=5.8$, then $\Gamma(6.8)<497$ whereas $5.8\times 4.8\times\ldots>533$). – Arnaud Mortier Dec 5 '19 at 13:37
• Ah yes, I didn’t notice that n was an integer. Then the bound would simply be ceil(a)! And floor(b)! then, no? – Gabe Dec 5 '19 at 13:54

There are two natural things to say:

• It's between $$\Gamma(a+1)$$ and $$\Gamma(b+1)$$ where $$\Gamma$$ is defined here.
• It's between $$a\cdot (a-1)\cdot\ldots\,$$ (stop when you reach a factor that is less than $$2$$) and $$b\cdot (b-1)\cdot\ldots\,$$ (same).

In fact it seems$$^{\color{blue}{\left[\underline{\text{reference needed}}\right]}^{\star}}$$ that $$\Gamma(x+1)$$ is always less than or equal to $$x\times (x-1)\times\ldots$$, meaning that the tightest couple of bounds would be $$a\cdot(a-1)\cdot\ldots\cdot (a-\lfloor a\rfloor+1)

$$^\star$$Edit: Peter Foreman proved the inequality $$\Gamma(x+1)\leq \prod_{k=0}^{\lfloor x\rfloor-1}(x-k)$$ in the comments below.

• Using the recursive defintion of $\Gamma(x+1)$ we have$$\Gamma(x+1)=\Gamma(x-\lfloor x\rfloor+1)\prod_{k=0}^{\lfloor x\rfloor-1}(x-k)$$but, as $1\le x-\lfloor x\rfloor+1\lt2$, we have $0.885603\lt\Gamma(x-\lfloor x\rfloor+1)\le1$ which implies that$$\Gamma(x+1)\le\prod_{k=0}^{\lfloor x\rfloor-1}(x-k)$$ – Peter Foreman Dec 5 '19 at 13:55
• @PeterForeman Nice. I'll include this in the body of the answer. – Arnaud Mortier Dec 5 '19 at 13:58

Similar to what Aurnaud Mortier said , I think the answer is that $$\lceil a \rceil !\le n! \le \lfloor b \rfloor !$$ , where $$\lceil a \rceil$$ is the ceiling function of $$a$$ and $$\lfloor b \rfloor$$ is the floor function of b. If $$a$$ and $$b$$ are integers, it's obvious why that is. If one of them is a real number, you can only have a factorial of a natural number, and you check the one imediatly above or down.

This is slighty more precise than between $$a⋅(a−1)\cdot{...}$$ and $$b⋅(b−1)\cdot{...}$$ , at least if you are looking for factorial for naturals. If you want to "extend" it to the reals, the gama function solution is the right one, this is for n as a positive integer.

*Edit

This is tighter than other solutions because for all $$x$$:

$$\lfloor x \rfloor ! \le Γ(x+1)$$