Answer
Converges
Work Step by Step
Let $u_n=\dfrac{5n^3-3n}{n^2(n-2) (n^2+5)}$ and $v_n=\dfrac{1}{ n^2}$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{5n^3-3n}{n^2(n-2) (n^2+5)}}{1/n^2}$
$\implies \lim\limits_{n \to \infty} \dfrac{(5/n-3/n^2)}{(1/n-2/n^2) (1+5/n^2)}=5$
Thus, the series converges by the limit comparison test.