terms $(x^n/n!)$ approach 0 faster than $(1/x)^n$

Question

Can anyone help me prove that the terms $\frac{x^n}{n!}$ approach $0$ faster than $(\frac{1}{x})^n$. Thank you in advance for the help.

Answer 1

If you want to prove that $a_n$ approaches $0$ faster than $b_n$, you usually need to prove that $$\lim_{n\to\infty}\frac{a_n}{b_n}=0.$$ In your case, you want to show that $$\lim_{n\to\infty}\frac{x^n\cdot x^n}{n!} = 0.$$

This is no more difficult than proving that $\frac{x^n}{n!}$ has a limit of $0$ for any value of $x$, since $$\frac{x^n\cdot x^n}{n!} = \frac{(x^2)^n}{n!},$$ therefore the arguments proving $\frac{x^n}{n!}$ converges to zero will also hold in your case.