real analysis - Prove that for $a > 0$, $lim_{n rightarrow infty}{frac{a^n}{n!}=0}$

Prove that for $a > 0$, $\lim_{n \rightarrow \infty}{\frac{a^n}{n!}=0}$. My attempt is since

$$e^x=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$ and the series converges for all $x \in \mathbb{R}$, by the test of divergence, $$\lim_{n \rightarrow \infty}{\frac{a^n}{n!}=0}.$$ Is my proof correct ? or are there any alternative ?

Answer

In the interest of providing a (hopefully intuitive) alternative, since you asked for it:

By expanding the factorial (and thereby hopefully making it more tangible) it is fairly easy to see that it grows quicker than the exponential.

$$\frac{a^n}{n!} = \frac{a}{1}\frac{a}{2}\frac{a}{3}\ldots\frac{a}{n}$$

The exponential is simply the repeated product of $n$ number of $a$'s, whereas the factorial is the repeated product of $n$ ever-growing integers.

Therefore there must come a point, some $n \ge a$, after which all new factors of the denominator must be larger than the new factors of the numerator. This gap in the size of $a$ and $n$ will only continue to grow; the final factor itself, $\frac{a}{n}$, goes to $0$ as $n$ does, and so the product of an infinite amount of such small fractions has little choice but to follow suit and approach $0$.