## Calculus: Early Transcendentals 8th Edition

(a) To determine whether $a_n$ converges, take the limit of $a_n$ as $n$ approaches $\infty$: $\lim\limits_{n \to \infty}a_n = \lim\limits_{n \to \infty}\frac{2n}{3n+1}=\frac{2}{3}$ since the numerator and denominator are of the same degree. Since $\lim\limits_{n \to \infty}a_n$ exists, $\{a_n\}$ is convergent. (b) From part (a), $\lim\limits_{n \to \infty}a_n =\frac{2}{3}$. By the Test for Divergence, since $\lim\limits_{n \to \infty}a_n \ne0$, the series $\Sigma_{n=1}^{\infty}a_n$ must diverge.