## University Calculus: Early Transcendentals (3rd Edition)

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ Thus, we have $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} \ln \dfrac{1}{3^n}=-\infty \ne 0$ Now, $s_n \to \infty$ as $n \to -\infty$, so, the given series diverges.