proof of limit involving factorials and exponents

Question

$ \cdot \lim \limits_{n \to \infty}\frac{10^n}{n!} $

I know intuitively that this is zero but I'm not sure how to prove this.

Can I use an inequality? Maybe $\frac{10^n}{n!} \le \frac{1}{n!}$ when n is large but how large?

Maybe I can take the $ln(\frac{10^n}{n!}) = \frac{nln(10)}{ln(n!)} $

I'm not sure

Answer 1

Hint: What happens when n > 10.

10/11 * 10/12 * ... * 10/100000 * 10/100001*...

What do you notice about the terms of the above product.

Answer 2

hint:$\dfrac{10^n}{n!} \leq \dfrac{1}{n}, n > 100$

Answer 3

$$ \frac{10^n}{n!} = \frac{10^{10}}{10!} \cdot \frac{10}{11} \cdots \frac{10}{n} < \frac{10^{10}}{10!} \frac{10}{n}$$

As $n$ approaches $\infty$, the above fraction approaches $0$.

Answer 4

We can apply the ratio test to the sequence $a_n = \frac{10^n}{n!}$:

$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{10^{n+1}\cdot n!}{10^{n}\cdot (n+1)!}=\lim_{n\to\infty}\frac{10}{n+1}=0\color{red}{<1}\implies \lim_{n\to\infty}a_n = 0$$