Prove that factorial n is less than or equal to n raised to n

issacnewton

Member
Hello

I wish to prove that
$\forall\;n\in \mathbb{N}\; n! \leqslant n^n$
First we let $$n$$ be arbitrary. Now I first write $$n!$$ as $$n\cdot(n-1)\cdot(n-2)\cdots 3\cdot 2\cdot 1$$.
Now we see that
$n \geqslant (n-1)\;; n \geqslant (n-2)\;\ldots ;n \geqslant n- (n-1)$
So we get
$n\cdot(n-1)\cdot(n-2)\cdots 3\cdot 2\cdot 1 \leqslant \underbrace{n\cdot n\cdot n\cdots n}_\text{n times}$
$\Rightarrow n! \leqslant n^n$
Since $$n$$ is arbitrary, the result is generally true. I want to use this result to find the limit of $$(n!)^{1/n^2 }$$
using the Squeeze theorem. So is my proof correct ?
Thanks

ZaidAlyafey

Well-known member
MHB Math Helper
Hello

I wish to prove that
$\forall\;n\in \mathbb{N}\; n! \leqslant n^n$
First we let $$n$$ be arbitrary. Now I first write $$n!$$ as $$n\cdot(n-1)\cdot(n-2)\cdots 3\cdot 2\cdot 1$$.
Now we see that
$n \geqslant (n-1)\;; n \geqslant (n-2)\;\ldots ;n \geqslant n- (n-1)$
So we get
$n\cdot(n-1)\cdot(n-2)\cdots 3\cdot 2\cdot 1 \leqslant \underbrace{n\cdot n\cdot n\cdots n}_\text{n times}$
$\Rightarrow n! \leqslant n^n$
Since $$n$$ is arbitrary, the result is generally true. I want to use this result to find the limit of $$(n!)^{1/n^2 }$$
using the Squeeze theorem. So is my proof correct ?
Thanks
Yup , it is correct .

issacnewton

Member
Thanks zaid.
Now to use squeeze theorem, one of the things I need to prove is that $$(n!)^{1/n^2} \leqslant (n^n)^{1/n^2}$$. Now here is an idea how I plan to go about doing that.
I have already proven a theorem that if $$a>0$$ and $$b>0$$ and $$n\in\mathbb{N}$$, we have $$a< b$$ if and only if $$a^n < b^n$$. Since $$n! \leqslant n^n$$ for all $$n\in\mathbb{N}$$, we can divide this in two cases. In case 1 , where $$n! = n^n$$,
taking $$1/n^2$$ root of both the sides, we get $$(n!)^{1/n^2} = (n^n)^{1/n^2}$$.
In case 2, where $$n! < n^n$$, I will let $$a = (n!)^{1/n^2}$$ and $$b = (n^n)^{1/n^2}$$. Now since both $$n!$$ and $$n^n$$ are positive, by $$n^{\mbox{th}}$$ root theorem, we have $$a = (n!)^{1/n^2} > 0$$ and
$$b = (n^n)^{1/n^2} > 0$$. The case 2 says that $$n! < n^n$$, which is $$a^{n^2} < b^{n^2}$$. Now since $$n\in\mathbb{N}$$, we have $$n^2 \in\mathbb{N}$$. So using the theorem I proved, we get $$a<b$$ which is $$(n!)^{1/n^2} < (n^n)^{1/n^2}$$. So combining the two cases, I get
$(n!)^{1/n^2} \leqslant (n^n)^{1/n^2}$.

Does this sound good ?

thanks

ZaidAlyafey

Well-known member
MHB Math Helper
It is clear that you know what you are doing but you need to make things a little be more organized . For example , it is not clear from the context which implication you are choosing ! .

Let $$\displaystyle a,b>0$$ , $$\displaystyle a<b \iff a^{n^2}<b^{n^2}$$ . so you are choosing the converse $$\displaystyle a^{n^2}<b^{n^2} \implies a< b$$ then by letting $$\displaystyle a= (n!)^{\frac{1}{n^2}} ,b= (n^n)^{\frac{1}{n^2}}$$ , we conclude that that $$\displaystyle n! < n^n \implies (n!)^{\frac{1}{n^2}} < (n^n)^{\frac{1}{n^2}}$$ .Since you already proved the left hand side we are done .

issacnewton

Member
Thanks.........I am on right track........