## Thomas' Calculus 13th Edition

Let $u_n=\dfrac{( n-2)}{n^3-n^2+3}$ and $v_n=\dfrac{ 1}{n^2}$ Then $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{(n-2)}{(n^3-n^2+3)}}{\dfrac{ 1}{n^2}}$ $\implies \lim\limits_{n \to \infty} \dfrac{n^3-2n^2}{n^3-n^2+3}=1$ Thus, the series converges due to the limit comparison test.