TITLE: How to prove that $\lim\limits_{n\to\infty} \frac{n!}{n^2}$ diverges to infinity?
QUESTION [5 upvotes]: $\lim\limits_{n\to\infty} \dfrac{n!}{n^2} \rightarrow \lim\limits_{n\to\infty}\dfrac{\left(n-1\right)!}{n}$
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?

REPLY [2 votes]: 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 $(n+1)(n+1)! - n n! = n!((n+1)^2-n) = (2n+1)n!$,
$$\lim_{n \to \infty}\dfrac{n!}{n^2}
= \lim_{n \to \infty}\dfrac{n n!}{2n+1}
= \lim_{n \to \infty}\dfrac{(2n+1) n!}{2}
\to \infty
$$