$$ x_n = \frac {1} {\sqrt[n] {n!}} $$
My attempt is to prove that $n!$ is always larger than the $a^n$ (with $a$ as a given constant, a as large as possible) . So the limit would be like:
$$ 0<\lim_{n \rightarrow \infty} x_n< \frac {1} {a} $$
And as $a$ is as large as possible, $\frac {1} {a}$ approaches 0. So the $\lim _{n \rightarrow \infty} x_n$ would approaches 0.
I kinda got the result I want but wonder if there’s a more accurate way to present this? Thanks a lot! Sorry for my kinda rough English because we do not study Calculus in English
