Answer
converges to 0.
Work Step by Step
$n^{2}e^{-n}=\frac{n^{2}}{e^{n}}$
Using L'Hopital's Rule (taking the derivative of the numerator and denominator of the limit)
$\lim\limits_{n \to \infty}\frac{n^{2}}{e^{n}}=\lim\limits_{n \to \infty}\frac{2n}{e^{n}}=\lim\limits_{n \to \infty}\frac{2}{e^{n}}=0$
Therefore, the sequence converges to 0.