Inequality (Factorial Related)

**anirban** · August 2nd 2008, 10:16 AM

Prove : $(n!)^2 > n^n$ when n is a +ve integer > 2

**flyingsquirrel** · August 3rd 2008, 12:28 AM

Hi !

Originally Posted by anirban

Prove : $(n!)^2 > n^n$ when n is a +ve integer > 2

$n! = 1\times 2\times 3\cdots (n-1)\times n$ so

$\begin{aligned}(n!)^2 &= [1\times 2\times 3\cdots (n-1)\times n]^2\\<br /> &={\color{red}1}\times {\color{blue}1} \times {\color{green}2}\times {\color{magenta}2} \cdots {\color{magenta}(n-1)}\times{\color{green}(n-1)}\times {\color{blue}n}\times {\color{red}n}\end{aligned}<br />$

Each term in this product appears twice so we can rewrite it this way

$\begin{aligned}(n!)^2 &= \underbrace{\color{red}[1\cdot n]}_{1^{\text{st}}\text{ term}} \times \underbrace{\color{green}[2\cdot (n-1)]}_{2^{\text{nd}}\text{ term}} \cdots \underbrace{\color{magenta}[(n-1)\cdot 2]}_{(n-1)^{\text{th}}\text{ term}}\times \underbrace{\color{blue}[ n\cdot 1]}_{n^{\text{th}}\text{ term}}\end{aligned}<br />$

We want to show that $(n!)^2>n^n$ that is $\frac{(n!)^2}{n^n}>1$ for $n>2$ . As $n^n=\underbrace{n\times n \cdots n \times n}_{n\text{ terms}}$ also contains $n$ terms, one can write

$\begin{aligned}\frac{(n!)^2}{n^n} &= \frac{1\cdot n}{n} \times \frac{2\cdot (n-1)}{n} \cdots \frac{(n-1)\cdot 2}{n}\times \frac{ n\cdot 1}{n}\end{aligned}<br />$

Now I claim that each fraction which appears in this product is greater than or equal to 1 and that for $n>2$ , there is at least one fraction which is $>1$ . If you manage to show this, the conclusion follows. Good luck !

**anirban** · August 3rd 2008, 01:32 AM

Thanks a lot! I felt as if I am looking at some solution book! Great illustration!

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