Show that $e^{-n}=o(\frac{1}{n^2})$

Question

Show that $e^{-n}=o(\frac{1}{n^2})$

I have to prove that $\lim_{n\to \infty}\frac{e^{-n}}{\frac{1}{n^2}}=0$, is to say that $\lim_{n\to \infty}\frac{n^2}{e^n}=0$, but I do not know how to do this, could someone give me a hint please? Thank you very much.

Answer 1

It suffices to show $e^n>n^3$ for $n$ big enough. But $$e^n=\sum_{k=0}^\infty \frac{n^k}{k!}>\frac{n^4}{24}>n^3$$ if $n>24$.

Answer 2

Hint: For all $n$

$$e^n = \sum_{k=0}^\infty \frac{n^k}{k!} > \frac{n^3}{3!} \implies 0 < \frac{n^2}{e^n} < \ldots$$