Answer
Converges
Work Step by Step
Let $u_n=\dfrac{(n-1)!}{(n+2) !}=\dfrac{(n-1)!}{(n+2) !}=\dfrac{1}{n(n+1)(n+2)}$ and $v_n=\dfrac{1}{ n^3}$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{1}{n(n+1)(n+2)}}{1/n^3}$
$ \implies \lim\limits_{n \to \infty} \dfrac{1}{(1+1/n)(1+2/n)}= 1 \ne 0 \ne \infty $
Here, $v_n=\Sigma_{n=1}^\infty \dfrac{1}{n^{3}}$ shows a convergent $p$-series with $p=3 \gt 1$
Hence, the series converges by the limit comparison test.
