# Using the sandwhich theorem to evaluate $\lim_{n \to \infty} \frac{n!}{n^n}$

In a question I am asked to find $\lim_{n \to \infty}\frac{n!}{n^n}$ using the sandwich theorem.

I can see that the limit is going to be 0 as $\frac{n!}{n^n}$ can be written as $\frac{n}{n}*\frac{n-1}{n}*\frac{n-2}{n}*...*\frac{1}{n}$ but I cannot see how I can find two sequences which 'squeeze' the given function.

Any help would be greatly appreciated!

## 3 Answers

We have $$0 < \frac{n!}{n^n} = \frac{n}{n} \frac{n-1}{n} \cdots \frac{1}{n} \leqslant 1 \cdot 1 \cdots 1 \cdot \frac{1}{n} = \frac{1}{n}$$ since the denominator of each term is bigger than the numerator. Now squeeze.

• Thanks very much! I hate how these questions are so simple once you see a possible solution! – P Collier Jul 30 '16 at 16:50

An idea: You can write: $n! = n(n-1)(n-2)\dots\frac{n}{2}\dots 1$. From there, $0\leq n! \leq n^{\frac{n}{2}-1}\cdot \left(\frac{n}{2}\right)^{\frac{n}{2}}$.

Can you conclude from there? (As a small remark: this will give you that the convergence to $0$ is exponentially fast.)

(Note that for convenience, i assumed above $n$ was even; if $n$ is odd, basically the same thing will work. You can add floors or ceilings to handle the two cases at once.)

• This is most likely not in your toolbox yet, but if you are interested in the exact rate of convergence, Stirling's approximation is the right "hammer." Namely, $\frac{n!}{n^n}$ behaves exactly like ${\sqrt{2\pi n}}e^{-n}$ when $n\to\infty$. – Clement C. Jul 30 '16 at 16:51
• Thanks for your answer! and yes I haven't come across that but I'll have a look now :) – P Collier Jul 30 '16 at 16:53
• You're welcome! Zain Patel's answer is simpler, if you don't care about showing that it goes to zero "very fast." (I would also encourage you to keep Stirling's in mind, for later -- it is a useful string to have to one's bow.) – Clement C. Jul 30 '16 at 16:54
• @ClementC. I posted a solution that refers to another solution I posted a few days ago in which I used the trapezoidal rule to show the behavior of $n!$ for large $n$. I did not develop the $\sqrt{2\pi}$ part which is, I believe, the more challenging piece of Stirling's formula. – Mark Viola Jul 30 '16 at 17:09
• Yes, pinpointing the constant is the "hardest" bit (without the constant, it becomes a much simpler exercise of real analysis). – Clement C. Jul 30 '16 at 17:20

PRIMER:

In THIS ANSWER, I showed using only straightforward arithmetic along with the Trapezoidal Rule that

$$n!=\sqrt{n}\left(\frac ne\right)^n\,O\left(1\right) \tag 1$$

Therefore, we see immediately that

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

as expected!