## Calculus: Early Transcendentals 8th Edition

If a series is convergent, then $\lim\limits_{n \to \infty}a_{n}=0$ $\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}ln(\frac{n}{3n+1})$ $=ln\frac{1}{3}\ne 0$ Hence, the series is divergent since the limit does not equal $0$, by the Test for Divergence.