Answer
Converges
Work Step by Step
Consider $a_n=n^3 e^{-n}$
Now, $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{(n+1)^3 e^{-(n+1)}}{n^3 e^{-n}}|$
Thus, we have $l=(\dfrac{1}{e})\lim\limits_{n \to \infty}(\dfrac{n+1}{n})^3=(\dfrac{1}{e})(1)$
So, $l=\dfrac{1}{e} \lt 1$
Hence, the series Converges by the ratio test.