How to calculate $\lim_{n\to \infty } \frac{n^n}{n!^2}$? 
How to calculate $\lim_{n\to \infty } \frac{n^n}{n!^2}$?


I tried to solve this by using Stirling's formula but I got stuck at $\lim\limits_{n\to \infty } \frac{e^{2n}}{n^{n+1}}$. Any help?
 A: Most of the time you do not need the exact Stirling approximation, just use the simplest: 
Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$
It ensues that $\dfrac{(n!)^2}{n^n}>\left(\dfrac n9\right)^n\to\infty$ so the inverse is going to $0$.

Anyway regarding the point where you are stuck $\dfrac{n^{n+1}}{e^{2n}}=\underbrace{n}_{\to\infty}\bigg(\ \underbrace{\dfrac n{e^2}}_{\to\infty}\ \bigg)^{\underbrace{n}_{\to\infty}}\to \infty$ 
so the inverse is going to $0$.
A: Alternative approach. Let $a_n=\frac{n^n} {n! ^2} $. It is clear that $a_n$ is nonnegative. Moreover we have \begin{align} \frac{a_{n+1} } {a_n} &=\frac{(n+1)^{n+1}n!^2}{n^n(n+1)!^2}\\ &=\frac{1}{(n+1)}\left(1+\frac{1}{n}\right)^n\to 0\end{align} 
Hence $\sum_{n\geq 1} a_n$  converges by the ratio test and that implies $\lim_{n\to \infty} a_n=0$. 
A: Go on!
$$
\lim_{n\to\infty}\frac{e^{2n}}{n^{n+1}}=\lim_{n\to\infty}\frac{1}{n}\left(\frac{e^2}{n}\right)^n
$$
For $n>18$ you have
$$
\frac{e^2}{n}<\frac{1}{2}
$$
Alternatively, consider the ratio
$$
\frac{(n+1)^{n+1}/((n+1)!)^2}{n^n/(n!)^2}=
\left(\frac{n!}{(n+1)!}\right)^{\!2}\left(\frac{n+1}{n}\right)^{\!n}(n+1)
=\frac{1}{n+1}\left(1+\frac{1}{n}\right)^{\!n}\to0
$$
Therefore the series
$$
\sum_{n\ge1}\frac{n^n}{(n!)^2}
$$
is convergent by the ratio test, in particular its general term has limit $0$.
A: Note that
$$a_n=\frac{n^n}{n!^2}=\frac{n^n}{(n+1)^n}\frac{(n+1)!^2}{n!^2(n+1)}\frac{(n+1)^{n+1}}{(n+1)!^2}=\frac{n+1}{\left({1+\frac1n}\right)^n}\frac{(n+1)^{n+1}}{(n+1)!^2}\geq\frac{n+1}{e}a_{n+1}>\frac{n}{e}a_{n+1}$$
thus $$a_{n+1}<\frac{e}{n}a_n$$
and $$a_1=1\implies a_2<\frac{e}{1}\implies a_3<\frac{e}{2}\frac{e}{1}\implies...\implies a_{n+1}<\frac{e^n}{n!}\to0 $$
thus for squeeze theorem
$$0\leq a_{n+1} <\frac{e^n}{n!}\implies \lim\limits_{n\to \infty } \frac{e^{2n}}{n^{n+1}}=0$$
