How to prove that $\lim_{n\to\infty}\frac{a^n}{n!}=0$?

Question

What I have got:

For any $a\in\mathbb{R}$, we can find an $N$, $N\gt a$, such that, $\lim_{n\to\infty}\frac{a^n}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdot\ldots\cdot\frac{a}{N-1}\cdot\frac{a}{N}\cdot\frac{a}{N+1}\cdot\ldots,$ and the later part $\frac{a}{N}\cdot\frac{a}{N+1}\cdot\ldots\lt\frac{a^m}{n^m}\to 0$ when $m\to\infty$, because $\frac{a}{n}\lt1$.

Then use the definition of convergence, with letting $M=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdot\ldots\cdot\frac{a}{N-1}$, and $\frac{a^m}{n^m}\lt\delta$ for $\delta=\frac{\epsilon}{M}$.

For an n that is large enough, we have $\frac{a^n}{n!}\lt\epsilon$, $\forall\epsilon\gt 0$.

What I want to ask is:

Are there any other ways to prove that this limit tends to $0$?

Thanks!

Answer 1

We have for $a\neq 0$: $$\log\frac{|a|^n}{n!}=n\log| a|-\sum_{k=1}^n\log k\sim_\infty n\log|a|-\int_1^n\log x dx=n\log|a|-n\log n\to-\infty$$ hence $$\lim_{n\to\infty}\frac{a^n}{n!}=0$$

Answer 2

You can use the result, if $ \lim_{n\to \infty} \frac{b_{n+1}}{b_n}=b $ and $|b|<1$, then $\lim_{n\to \infty} b_n = 0$. Applying this to your problem, we have

$$ \lim_{n \to \infty} \frac{a^{n+1}}{(n+1)!}\frac{n!}{a^n}=\lim_{n\to \infty}\frac{a}{n+1}=0 <1 $$

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

Answer 3

If some asks a reasonable question is better to answer it straightforward than to elaborate into unnecessary details. Use ratio test because it is more easier: $\lim_{n\to\infty}\frac{a^{n+1}}{{(n+1)}!}/\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{n+1}=0<1$. Hence $\lim_{n\to\infty}\frac{a^n}{n!}=0$. user66607 I hope you get it now.