Answer
converges
Work Step by Step
In order to solve the given series we will take the help of Ratio Test. This test states that when the limit $L \lt 1$, the series converges and for $L \gt 1$, the series diverges.
Let us consider $a_n=(n^2) e^{(-n)}$
Then, $L=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n+1}} |=\lim\limits_{n \to \infty}|\dfrac{(n+1)^2 e^{-(n+1)}}{n^2 e^{-n}}|$
$\implies (\dfrac{1}{e})\lim\limits_{n \to \infty}(\dfrac{(n+1)}{n})^{2}=(\dfrac{1}{e})(1)=\dfrac{1}{e} \lt 1$
Hence, the series Converges by the ratio test.