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?

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 $\lim_{n\to \infty} a_n=0$. 
