Show that $n^n<(n!)^2$

Question

I want to show that $\lim\limits_{n \to \infty}\frac{n^n}{(n!)^2}=0$

But I have absolutely no idea besides that $\frac{n^n}{(n!)^2}=\frac{n}{1}\cdot \frac{n}{2}\cdot ...\cdot \frac{n}{(n-1)^2}\cdot \frac{n}{n^2}$

Help me please.

Answer 1

Let's check ratio of $a_n$ and $a_{n+1}$: $$a_n = \frac{n^n}{n!^2}$$ $$a_{n+1} = \frac{(n+1)^{n+1}}{(n+1)!^2}$$ $$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(n+1)!^2} : \frac{n^n}{n!^2} = \frac{(n+1)^{n+1}}{n^n}\frac{n!^2}{(n+1)!^2} = \left(1+\frac{1}{n}\right)^n \frac{n!^2(n+1)}{n!^2(n+1)^2} = \frac{1}{n+1}\left(1+\frac{1}{n}\right)^n \sim\frac en, $$ hence $$a_n \sim \frac{e^n}{n!}\to 0.$$

Answer 2

Taking logarithms and using formula for arithmetic sum: $$n \log(n) - 2 \sum_{i=1}^n i = n\log(n) - 2n(n-1)/2 = n(\log(n) - n-1)$$This obviously goes to $-\infty$ since any polynomial dominates a logarithm and if the logarithm of something goes to $-\infty$, itself it must go to 0.

Answer 3

Let $a_{n}=\frac{n^{n}}{\left( n!\right) ^{2}}.$ We compute , as Michael Galuza did \begin{equation*} \frac{a_{n+1}}{a_{n}}=\frac{(1+\frac{1}{n})^{n}}{n+1}. \end{equation*} Note that for large $n,$ \begin{equation*} \frac{a_{n+1}}{a_{n}}<1, \end{equation*} so the positive sequence $(a_{n})$ is decreasing and bounded from below by $0.$ Then there exists $l\geq 0$ such that $\lim_{n\rightarrow \infty }a_{n}=l.$ ($(a_{n})$ converges to $l$). Assume that $l>0.$ Since $(a_{n+1})$ is a subsequence of $($a$_{n})$ it converges to $l$ too. Then \begin{equation*} \lim_{n\rightarrow \infty }\frac{a_{n+1}}{a_{n}}=\frac{l}{l}=1. \end{equation*} However \begin{equation*} \lim_{n\rightarrow \infty }\frac{a_{n+1}}{a_{n}}=\lim_{n\rightarrow +\infty }% \frac{1}{n+1}(1+\frac{1}{n})^{n}=0\cdot e=0. \end{equation*}