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$\lim\limits_{n\to\infty} \dfrac{n!}{n^2} \rightarrow \lim\limits_{n\to\infty}\dfrac{\left(n-1\right)!}{n}$

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?

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$$(n-3)(n-3)!<(n-3+\frac{2}{n})(n-3)!=\frac{n!}{n^2}$$ – Ethan Mar 8 '13 at 9:45
@Ethan how can $\left(n-3\right)=\left(n-3+\frac{2}{n}\right)$? – yiyi Mar 9 '13 at 4:58
Its an inequality not an equality – Ethan Mar 9 '13 at 5:53
up vote 11 down vote accepted

you could introduce the gamma function!

Just a joke, as $n!>n^3$ for $n>100$ you know that $$\frac{n!}{n^2} > \frac{n^3}{n^2}=n$$

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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 $\frac{n^2}{n!} \to 0$.

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By Stirling formula $$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ the result that you look for is clear.

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But don't you think the @Dominici's is faster cause the OP will be involved in bad figured limit next when he/she use your hint? – Babak S. Mar 8 '13 at 9:27
I think stirling's formula is overkill here.. – Ethan Mar 8 '13 at 9:40
@Ethan, have bad habits, every time I see $n!$ I think to Stirling. – user63181 Mar 8 '13 at 9:44

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.

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I agree with your statement, but wanted to show that the last step diverges. Why is it not infinity/infinity? – yiyi Mar 9 '13 at 5:02
$ \underset{n \rightarrow \infty}{\lim} \frac{n^2}{(n+1)} = \underset{n \rightarrow \infty}{\lim} \frac{\frac{1}{n}n^2}{\frac{1}{n}(n+1)} = \underset{n \rightarrow \infty}{\lim} \frac{n}{1+\frac{1}{n}} = \underset{n \rightarrow \infty}{\lim} \frac{1}{1+\frac{1}{n}}\cdot \underset{n \rightarrow \infty}{\lim} n = 1\cdot \infty = \infty$ – sponsoredwalk Mar 9 '13 at 6:28

$\dfrac{n!}{n^2}=(1-\frac 1n)(n-2)!$ 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.

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I don't know how you can expand $(1-\frac 1n)(n-2)!$ – yiyi Mar 9 '13 at 5:00
The first factor tends to one, the 2nd diverges, so the whole thing diverges. – Ishan Banerjee Mar 9 '13 at 6:24

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.


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 $(n+1)(n+1)! - n n! = n!((n+1)^2-n) = (2n+1)n!$, $$\lim_{n \to \infty}\dfrac{n!}{n^2} = \lim_{n \to \infty}\dfrac{n n!}{2n+1} = \lim_{n \to \infty}\dfrac{(2n+1) n!}{2} \to \infty $$

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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 .$

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its called the ratio test. – yiyi Mar 24 '13 at 15:19

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

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