How to prove that $\lim\limits_{n\to\infty} \frac{n!}{n^2}$ diverges to infinity? $\lim\limits_{n\to\infty} \dfrac{n!}{n^2} \rightarrow \lim\limits_{n\to\infty}\dfrac{\left(n-1\right)!}{n}$
I can understand that this will go to infinity because the numerator grows faster.  
I am trying to apply L'Hôpital's rule to this; however, have not been able to figure out how to take the derivative of $\left(n-1\right)!$ 
So how does one take the derivative of a factorial?  
 A: Dominic Michaelis is the 'right' answer for such a simple problem. This is just to demonstrate a trick that is often helpful in showing limits going off to $\infty$. Consider $$\sum_{n=1}^{\infty} \frac{n^2}{n!}$$ By the ratio test this converges. So the terms $\frac{n^2}{n!} \to 0$.
A: By Stirling formula 
$$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$
the result that you look for is clear.
A: Notice L'Hopital's rule is a theorem that applies to continuous functions, thus if you want to apply L'Hopital's Rule to a sequence you'll have to invoke something like the Stolz–Cesàro theorem unless you want to start invoking logarithmic derivatives & Euler Mascheroni constants etc...
An easier method is to apply the sequence equivalent of D'Alembert's ratio test (Theorem V P147) without needing to resort to any series argument, something that generalizes Puzzled's 'trick'. If you're willing to accept the statement and proof given in the link then the following argument should be enough:
$\underset{n \rightarrow \infty}{\lim} \frac{f(n+1)}{f(n)} = \underset{n \rightarrow \infty}{\lim}  \frac{\frac{(n+1)!}{(n+1)^2}}{\frac{n!}{n^2}}  = \underset{n \rightarrow \infty}{\lim}  \frac{(n+1)!}{(n+1)^2}\frac{n^2}{n!} =  \underset{n \rightarrow \infty}{\lim}  \frac{(n+1)n^2}{(n+1)^2}  =  \underset{n \rightarrow \infty}{\lim}  \frac{n^2}{(n+1)} = \infty $
Thus we see the sequence diverges.
A: $\dfrac{n!}{n^2}=(1-\frac 1n)(n-2)!$ Taking a limit, we find that it diverges. If you really want to use L'hospitals,  you'd have to use the gamma function. Note that this is very annoying.
A: You cannot take the derivative of
the factorial function
unless you extend it to
the Gamma function.
The factorial function,
as used by the poster,
is defined only at the non-negative
integers.
L'Hôpital's rule cannot be applied.
You can take the difference
and use a discrete version of
L'Hôpital's rule,
in this form:
If
$f$ and $g$ are positive increasing unbounded functions
then
$
\lim_{n\to \infty} \dfrac{f(n)}{g(n)}
=\lim_{n\to \infty} \dfrac{f(n+1)-f(n)}{g(n+1)-g(n)}
$
if the right-hand limit exists;
if the right-hand limit is $\infty$,
so is the left-hand limit.
Proof:
Suppose 
$ \lim_{n\to \infty} \dfrac{f(n+1)-f(n)}{g(n+1)-g(n)}
= L$.
Then,
for all large enough $n$,
$$\left|\dfrac{f(n+1)-f(n)}{g(n+1)-g(n)}-L\right|
< \epsilon
$$
or
$$-\epsilon < \dfrac{f(n+1)-f(n)}{g(n+1)-g(n)}-L
< \epsilon
$$
or
$$-\epsilon(g(n+1)-g(n)) < f(n+1)-f(n)-L(g(n+1)-g(n))
< \epsilon(g(n+1)-g(n))
$$
Summing this from $M$ to $N-1$,
$$-\epsilon(g(N)-g(M)) < f(N)-f(M)-L(g(N)-g(M))
< \epsilon(g(N)-g(M))
$$
or
$$-\epsilon(g(N)-g(M))+f(M)-Lg(M) < f(N)-Lg(N)
< \epsilon(g(N)-g(M))+f(M)-Lg(M)
$$
or
$$-\epsilon+\dfrac{\epsilon g(M)+f(M)-Lg(M)}{g(N)}
< \dfrac{f(N)}{g(N)}-L
<\epsilon+\dfrac{\epsilon g(M)+f(M)-Lg(M)}{g(N)}
$$
Since $g(N)$ is unbounded,
$\dfrac{\epsilon g(M)+f(M)-Lg(M)}{g(N)}$
can be made as small as you want
for large enough $N$,
so that
$$\lim_{N \to \infty} \dfrac{f(N)}{g(N)}=L$$
Suppose 
$ \lim_{n\to \infty} \dfrac{f(n+1)-f(n)}{g(n+1)-g(n)}
= \infty$.
Then,
for any $L > 0$,
for all large enough $n$,
$$L < \dfrac{f(n+1)-f(n)}{g(n+1)-g(n)}
$$
or
$$L(g(n+1)-g(n)) < f(n+1)-f(n)
$$
Summing this from $M$ to $N-1$,
$$L(g(N)-g(M)) < f(N)-f(M)
$$
or
$$L(g(N)-g(M))+f(M) < f(N)
$$
or
$$L+\dfrac{f(M)-Lg(M)}{g(N)}
< \dfrac{f(N)}{g(N)}
$$
Since $g(N)$ is unbounded,
$\dfrac{f(M)-Lg(M)}{g(N)}$
can be made as small as you want
for large enough $N$,
so that
$\lim_{N \to \infty} \dfrac{f(N)}{g(N)}\ge L$.
Since we can choose $L$
as large as we want,
$\lim_{N \to \infty} \dfrac{f(N)}{g(N)}= \infty$.
Applying this twice to $\dfrac{n!}{n^2}$,
and using
$(n+1)!-n! = n n!$
and $(n+1)(n+1)! - n n! = n!((n+1)^2-n) = (2n+1)n!$,
$$\lim_{n \to \infty}\dfrac{n!}{n^2}
= \lim_{n \to \infty}\dfrac{n n!}{2n+1}
= \lim_{n \to \infty}\dfrac{(2n+1) n!}{2}
\to \infty
$$
A: you could introduce the gamma function!
Just a joke, as $n!>n^3$ for $n>100$ you know that 
$$\frac{n!}{n^2} > \frac{n^3}{n^2}=n$$
A: Here is a usful result you can use 

if $\lim_{n\to \infty} \frac{a_{n+1}}{a_n} = b $ and $|b|>1$, then $ \lim_{n\to \infty} a_n = \infty .$

A: If $n\geq 2$, then
$$ \frac{n!}{n^2} = \frac{n-1}{n}(n-2)! \geq \frac{1}{2}(n-2)!\rightarrow \infty \text{ as } n\rightarrow\infty.$$
