What's the limit of the sequence $\lim\limits_{n \to\infty} \frac{n!}{n^n}$?

Question

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

I have a question: is it valid to use Stirling's Formula to prove convergence of the sequence?

Answer 1

There are two distinct questions here. The first one in the title is what the limit actually is. This is easy to see by writing out the expression as a product of $n$ positive factors: $$\frac{n!}{n^n}=\left(\frac{1}{n}\right)\left(\frac{2}{n}\right)\left(\frac{3}{n}\right)\cdots\left(\frac{n}{n}\right).$$ Every one of the factors $k/n$, $k=1,2,3,\dots,n$, is less than or equal to $1$. Hence the product is $$\le\left(\frac{1}{n}\right)\cdot1\cdot1\cdots1=1/n.$$ But $1/n$ converges to $0$ as $n\to\infty$, so by the Squeeze theorem so does the original expression.

The second question is whether or not it's allowed to use Stirling's formula to derive the limit, which I believe Arturo's comment covers: there is no apparent circularity, but in the context of classwork the answer depends on whether or not you've formally learned the formula and are allowed to use it as a given.

Answer 2

No , both are equal in terms of their order of growth, obviously in terms of values $n^n$ is greater.

$n!=n(n-1)(n-2)(n-3).........1.$

Here , $n$ multiplies by $n$ times

So, $n!= n^n -$ something ,

here definitely something is less than $n^n$.

And in a function generally we considers highest order function among others , so $n! = n^n$ in terms of order of growth. It's simply like $f(n)= n^2+n+2 , g(n)= n^2 + 2$ , here both are qudartic as well as they have $n^2$ as highest order function , so they are equal in growth.

Answer 3

Let $u_n=\frac{n!}{n^n}$ then $u_{n+1}=\frac{(n+1)!}{(n+1)^{(n+1)}}$. Now, $$L=\lim_{n\to \infty}\frac{u_{n+1}}{u_n}\Rightarrow \lim_{n\to \infty}\frac{(n+1)!}{(n+1)^{(n+1)}}\times \frac{n^n}{n!}\\ =\lim_{n\to \infty}\left(\frac{n}{n+1}\right)^n=e^{-1}<1.$$ By ratio test for sequences the limit $\displaystyle \lim_{n\to \infty}\frac{n!}{n^n}=0.$

Answer 4

We will first show that the sequence $x_n = \frac{n!}{n^n}$ converges. To do this, we will show that the sequence is both monotonic and bounded.

Lemma 1: $x_n$ is monotonically decreasing.

Proof. We can see this with some simple algebra:

$$x_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}} = \frac{n+1}{n+1}\frac{n!}{(n+1)^n} \frac{n^n}{n^n} = \frac{n!}{n^n} \frac{n^n}{(n+1)^n} = x_n \big(\frac{n}{n+1}\big)^n.$$

Since $\big(\frac{n}{n+1}\big)^n < 1$, then $x_{n+1} < x_n$.

Lemma 2: $x_n$ is bounded.

Proof. Straightforward to see that $n! \leq n^n$ and $n! \geq 0$. We obtain the bounds $0 \leq x_n \leq 1$, demonstrating that $x_n$ is bounded.

Together, these two lemmas along with the monotone convergence theorem proves that the sequence converges.

Theorem: $x_n \to 0$ as $n \to \infty$.

Proof. Since $x_n$ converges, then let $s = \lim_{n \to \infty} x_n$, where $s \in \mathbb{R}$. Recall the relation in Lemma 1:

$$x_{n+1} = x_n \big(\frac{n}{n+1}\big)^n = \frac{x_n}{(1+ \frac{1}{n})^n}.$$

Since $x_n \to s$, then so does $x_{n+1}$. Furthermore, a standard result is the limit $(1+ \frac{1}{n})^n \to e$. With these results, we have $\frac{x_n}{(1+ \frac{1}{n})^n} \to \frac{s}{e}$ and consequently

$$s = \frac{s}{e} \implies s(1 - e^{-1}) = 0$$

Since $1 \neq e^{-1}$, then this statement is satisfied if and only if $s = 0$ and that concludes the proof.

Answer 5

My two cents: using $$\left(\frac{n}{e}\right)^n<n!<e\left(\frac{n}{2}\right)^n\tag{1}$$ is easy to obtain the estimation $$\frac{n!}{n^n}<e\cdot \frac{1}{2^n}\to 0,$$

I write down proof for right hand side of $(1)$, as we are using it, and left can be obtain similarly.

Using classical inequality for arithmetic and geometric averages $\frac{x_1+x_2+\cdots +x_n}{n}\geqslant \sqrt[n]{x_1x_2\cdots x_n}$ on first step, we have $$n!<\left(\frac{n+1}{n}\right)^n=e\left(\frac{n}{2}\right)^n\frac{\left(\frac{n+1}{n}\right)^n}{e\left(\frac{n}{2}\right)^n}=\\ =e\left(\frac{n}{2}\right)^n\cdot \frac{\left(1+\frac{1}{n}\right)^n}{e}<e\left(\frac{n}{2}\right)^n$$