## Calculus (3rd Edition)

$a_n$ converges to $-\frac{3}{2}$.
We have $$\lim _{n \rightarrow \infty} a_n=\lim _{n \rightarrow \infty} \frac{3n^3-n}{1-2n^3} \\ =\lim _{n \rightarrow \infty} \frac{3-(1/n^2)}{(1/n^3)-2 } =\frac{3-0}{0-2}=-\frac{3}{2}.$$ Hence, $a_n$ converges to $-\frac{3}{2}$.