Solution verification: showing $\lim \limits_{n \to \infty} \frac{n!}{n^n} = 0$

In a textbook, I came upon the following excercise:

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

I came up with the following solution:

\begin{align*} \lim_{n \to \infty} \frac{n!}{n^n} &= \lim_{n \to \infty} \prod_{i=1}^n \frac{n+1-i}{n} \\ &= \lim_{n \to \infty} \left(\prod_{i=1}^{n-1} \frac{n+1-i}{n} \cdot \frac{1}{n} \right) \\ &= \left( \lim_{n \to \infty} \prod_{i=1}^{n-1} \frac{n+1-i}{n} \right) \left( \lim_{n \to \infty} \frac{1}{n} \right) \end{align*} And since $$\lim_{n \to \infty} \frac{1}{n} = 0$$ we have $$\lim_{n \to \infty} \frac{n!}{n^n} = 0$$ as required.

Since my textbook gives another solution, I post mine here with the question, if this is correct. Thanks in advance for any feedback.

• Why not just show the term is between 0 and 1/n for any n? That should prove it. Apr 21, 2022 at 19:49
• Is it true that $$\lim 1=\lim_{n\to\infty}n \cdot \lim_{n\to\infty}\frac1n=0?$$ Apr 21, 2022 at 19:53
• Please don’t write it down like that. Do you think there’s difference between lim n and lim n^2 as n approaches infinity? Then according to your notation, surely lim n is 0 as n approaches infinity. That’s some misleading notation that’ll get you nowhere. Apr 21, 2022 at 20:01

As is, your proof doesn't work. The idea is good though, and can be made rigorous by studying the behavior of the product.

Write : $$\frac{n!}{n^n} = \frac 1n\prod_{i=1}^{n-1} \frac{n+1-i}{n}$$ Then, since each of the factor in the rightmost product is in $$[0,1]$$, so it the whole product and we have : $$0\leq \frac{n!}{n^n} \le \frac{1}{n}$$

Then, since $$\lim_{n\to +\infty} \frac{1}{n} = 0$$, by comparison we have : $$\lim_{n\to +\infty} \frac{n!}{n^n} = 0$$

This solution won't work, since it assumes that the other factor in the limit,

$$\lim_{n \to \infty} \prod_{i=1}^{n-1} \frac{n+1-i}{n}$$

exists and is finite in particular. What if it's infinite?

As a slightly sillier example, your logic suggests

$$\lim_{n \to \infty} n^2 = \left( \lim_{n \to \infty} n^3 \right) \left( \lim_{n \to \infty} \frac 1 n \right) = 0$$

but clearly the left-hand factor and the original limit are infinite.