Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{5n^3-3n}{n^2(n-2) (n^2+5)}$ and $b_n=\dfrac{1}{ n^2}$
Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{5n^3-3n}{n^2(n-2) (n^2+5)}}{\dfrac{1}{ n^2}}$
Thus, we have $ =\lim\limits_{n \to \infty} \dfrac{n^2(5n^3-3n)}{n^2(n-2) (n^2+5)}$
or, $=5$
Hence, the series converges due to the limit comparison test.
