Proof that $\lim\limits_{n\to \infty}\frac{n!}{n^n}$ is $0$

Solution 1:

It all looks good to me. It's difficult to address your concerns when I have no specific knowledge as to what's bothering you. Maybe presenting your proof in one line might help?

$$a_n = \frac{n}{n} \cdot \frac{n-1}{n} \cdot \ldots \cdot \frac{2}{n} \cdot \frac{1}{n} \le 1 \cdot 1 \cdot \ldots \cdot 1 \cdot \frac{1}{n} = \frac{1}{n} \to 0.$$

Perhaps your concern regards the expanding product, the number of terms of which increases with $n$? Maybe you're connecting it with the somewhat paradoxical nature of Riemann sums, adding $0$ "infinity times" to get a non-zero number? This isn't applicable here. Your proof shows that, at each step (when there's a finite number of terms), the total result of the product is bounded above by $\frac{1}{n}$. There's nothing wrong with the proof.

As for an alternate proof, you can show this is to show the infinite series $\sum_n a_n$ converges, by way of the ratio test. Consider, $$\frac{a_{n+1}}{a_n} = \frac{(n + 1)! n^n}{n! (n + 1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(1 + \frac{1}{n}\right)^{-n} \to \frac{1}{e} < 1.$$ By the ratio test $\sum_n a_n$ converges, and hence by the divergence test, $a_n \to 0$.