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

issacnewton

Member
Jan 30, 2012
61
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
Jan 17, 2013
1,667
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
Jan 30, 2012
61
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
Jan 17, 2013
1,667
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
Jan 30, 2012
61
Thanks.........I am on right track........ (Muscle)