Prove $\lim \limits_{n \to \infty}\frac{n!}{n^n}=0$

Question

Prove $\lim \limits_{n \to \infty}\frac{n!}{n^n}=0$. I have tried this several ways. I tried using the ratio test, I tried to expand it... I'm not really sure where to go.

When I expand, I get $$\frac{n(n-1)(n-2)(n-3).....}{nnnnnn......}$$ I don't know if trying to expand this is helpful, but I'm just not sure where to go with this problem... Any help would be appreciated.

$\frac{n!}{n^n} = \frac{1}{n}\prod_{k=2}^n\frac{k}{n} \le \frac{1}{n}$ — achille hui, Dec 14 '16 at 02:18
kinda irritating when teachers ask us to prove something that is so obviously true that just looking at it should be proof enough haha. — stack reader, Dec 14 '16 at 02:21
@stackreader Often, the purpose of proving an "obvious" statement is to practice using some important technique (approximating with upper bounds, in this case). — Austin Mohr, Dec 14 '16 at 02:45
See Compute the limit $\lim_{n \to \infty} \frac{n!}{n^n}$ and other posts linked there. — Martin Sleziak, Dec 14 '16 at 05:36

score 4 · Accepted Answer · answered Dec 14 '16 at 02:18

4

You've got the right idea. Note that $$\frac{n!}{n^n}=\frac{n}{n}\frac{n-1}{n}\cdots\frac{1}{n}$$ Every term in this product is at most $1$ so we have that $$\frac{n!}{n^n}=\frac{n}{n}\frac{n-1}{n}\cdots\frac{1}{n}\leq\frac{1}{n}$$

answered Dec 14 '16 at 02:18

user293794

3,668

Awesome! I like it. – setholopolus Dec 14 '16 at 02:44
Just an added bit of clarity: We are replacing every factor (except for the last one!) with its upper bound of $1$. – Austin Mohr Dec 14 '16 at 04:13

score 3 · Answer 2 · answered Dec 14 '16 at 02:29

3

Here is a combinatorial proof. The numerator is the number of bijections from $[n]\to[n]$; the denominator is the number of functions from $[n]\to [n]$. For each bijection $\sigma$, we can define $n$ different functions $\tau_1,\dots,\tau_n$ s.t. $\tau_i|_{[n-1]}=\sigma$ and $\tau_i(n)=i$. So the expression in the limit is $\leq 1/n$.

answered Dec 14 '16 at 02:29

TorsionSquid

3,530

1

(+1) Nice combinatorial interpretation. Just rephrasing: we want to estimate the probability that a random function $f:A\to A$, with $|A|=n$, is injective. Things may "go wrong" in so many ways that the limit probability is clearly zero. – Jack D'Aurizio Dec 14 '16 at 14:04

score 1 · Answer 3 · answered Dec 14 '16 at 02:37

1

The ratio test works:

$$\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^{n}}}=\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!}=\left(\frac{n}{n+1}\right)^n=\frac{1}{\left(1+\frac{1}{n}\right)^n}\overset{n\to\infty}{\longrightarrow} \frac{1}{e}<1.$$

answered Dec 14 '16 at 02:37

Pedro

18,817
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score 1 · Answer 4 · answered Dec 14 '16 at 03:38

Note that since $\sum_{k=1}^{n-1}\log(k)\le \int_1^{n} \log(x)\,dx$, we have

$$\begin{align} (n-1)!&=e^{\log((n-1)!)}\\\\ &=e^{\sum_{k=1}^{n-1}\log(k)}\\\\ &\le e^{\int_1^{n} \log(x)\,dx}\\\\ &=n^ne^{-(n+1)} \end{align}$$

Therefore, we see that

$$0\le \frac{n!}{n^n}\le ne^{-(n+1)}$$

whereupon application of the squeeze theorem yields the coveted limit

$$\lim_{n\to \infty}\frac{n!}{n^n}=0$$

And we are done!

Prove $\lim \limits_{n \to \infty}\frac{n!}{n^n}=0$

4 Answers4