If I have the limit $$\lim_{n\rightarrow\infty} \frac{n!}{n^2},$$ how do I prove that limit does not converge?
I tried to find two subsequence, but could not find them. I also tried Stirling's approximation for $n!$ and I obtained the limit $$\lim_{n\rightarrow\infty}\sqrt{2\pi}\frac{n^{n-\frac12}}{\mathrm e^n}$$ but I am stuck.
If you want to use Stirling's Formula, then note that
$$\frac{n^{n-1/2}}{e^n}=e^{-1/2}\left(\frac ne\right)^{n-1/2}$$
For $n>2e$, we have as $n\to \infty$
$$e^{-1/2}\left(\frac ne\right)^{n-1/2}>(2e)^{-1/2}(2)^n\to \infty$$
And we are done!
Notice that: $$ \lim_{n\to\infty}\frac{n^2}{n!}=\lim_{n\to\infty}\frac{1}{(n-2)!}\cdot\lim_{n\to\infty}\frac{n^2}{(n-1)n}=0\cdot 1=0. $$ Depending on whether one uses the extended real number $+\infty$, one may say that $\displaystyle \lim_{n\to\infty}\frac{n!}{n^2}=+\infty$ or the limit of the sequence $(\sqrt{n!}{n^2})_{n=1}^\infty$ does not exist.
