비교 판정법

For large $n$, the numerator $n^3-4\approx n^3$ and the denominator $2n^5-6n\approx 2n^5$, so the $n$th term is approximately $\frac{n^3}{2n^5}=\frac{1}{2n^2}$. So we apply the limit comparison test with $a_n=\frac{n^3-4}{2n^5-6n}$ and $b_n=\frac{1}{n^2}$. Since

$$\lim_{n\rightarrow\infty}\frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{(n^3-4)/(2n^5-6n)}{1/n^2} =\lim_{n\rightarrow\infty}\frac{1-\frac{4}{n^3}}{2-\frac{6}{n^4}} =\frac12$$

exists and is nonzero, the given series $\sum\limits_{n=1}^\infty a_n$ converges if and only if the series $\sum\limits_{n=1}^\infty b_n$ converges. Since the series $\sum\limits_{n=1}^\infty b_n =\sum\limits_{n=1}^\infty\frac{1}{n^2}$ is a convergent $p$-series (with $p=2$), both series converge.