Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{(n-1)!}{(n+2) !}$
or, $a_n=\dfrac{(n-1)!}{(n+2) !}=\dfrac{1}{n(n+1)(n+2)}$ and $b_n=\dfrac{1}{ n^3}$
Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{1}{n(n+1)(n+2)}}{\dfrac{1}{ n^3}}$
Thus, we have $ =\lim\limits_{n \to \infty} \dfrac{1}{(1+1/n)(1+\dfrac{2}{n})}$
or, $ \dfrac{1}{(1+0)(1+0)}=1 \ne 0 \ne \infty $
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{3}}$ is a convergent series due to the p-series test with $p=3 \gt 1$
Thus, the series converges by the limit comparison test.
