Answer
Converges to $0$
Work Step by Step
Given: $a_n=n^{2}e^{-n}$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}n^{2}e^{-n}$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{n^{2}}{e^{n}}$
Use L-Hospital's Rule.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{2n}{e^{n}}$
Again use L-Hospital's Rule.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{2}{e^{n}}$
$=\frac{2}{\infty}$
$=0$
Hence, the sequence converges to $0$.
