Answer
Converges
Work Step by Step
Let us consider $a_n=n^3 e^{-n}$
$\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{(n+1)^3 e^{-(n+1)}}{(n^3) (e^{-n})}|$
$\implies (\dfrac{1}{e})\lim\limits_{n \to \infty}(\dfrac{n+1}{n})^3=\dfrac{1}{e} \lt 1$
Therefore, the series converges by the ratio test.