Riley Hobson Mathematical Methods in Physics Page $831$
Since $(n!)^{1/n}$ behaves like $n$ as $n \rightarrow \infty$, we find $\lim(1/n!)^{1/n} =0$
Now I don't understand how does $(n!)^{1/n}$ behave like $n$ as $n \rightarrow \infty$.
I have seen the graph on Desmos and it seems like it becomes a straight line after $n=-1$. I don't understand this idea of it approaching $n$.
More precisely, $(n!)^{1/n}$ behave like $n/e$ where $e$ is Euler's constant. One way to see this is Stirling's approximation. This still implies that $1/[(n!)^{1/n}]$ approaches zero.
Stirling's formula: $n!\approx \sqrt{2\pi n}(\frac{n}{e})^n$ so $(n!)^{1/n}\approx \frac{n}{e}$.
Other people have already explained why this is true - but I thought I'd add a little more too.
Another version of Stirling's formula is the following: $$ \ln(n!) \sim n\ln(n) - n $$ what do I mean here by the ${\sim}$ symbol? It's the "asymptotic to" symbol. It means if you have two functions ${f(x),g(x)}$, we say that ${f\sim g}$ at infinity if $$ \lim_{x\to \infty}\frac{f(x)}{g(x)}=1 $$ you can see how this essentially tells us ${f(x)}$ and ${g(x)}$ behave the same at infinity. If we exponentiate both sides, we get $$ n! \sim \frac{n^n}{e^n} $$ and so now, $$ (n!)^{\frac{1}{n}}\sim \left(\frac{n^n}{e^n}\right)^{\frac{1}{n}} = \frac{n}{e} $$ i.e. $$ (n!)^{\frac{1}{n}}\sim \frac{n}{e} $$ this tells you at infinity, ${(n!)^{\frac{1}{n}}}$ behaves like ${\frac{n}{e}}$. A straight line with gradient ${\frac{1}{e}}$.
Asymptotic relations are helpful as they allow you to deduce things like the above, and also help you in calculating limits. See my answers in the following posts to see how: Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero and Is this approach to the limit solution kosher?$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$
