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