TITLE: Show that $e^{-n}=o(\frac{1}{n^2})$
QUESTION [1 upvotes]: 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.

REPLY [2 votes]: 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 $n>24$.

REPLY [1 votes]: Hint: For all $n$
$$e^n = \sum_{k=0}^\infty \frac{n^k}{k!} > \frac{n^3}{3!} \implies 0 < \frac{n^2}{e^n} < \ldots$$
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