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

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ Thus, we have $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} \ln ( \dfrac{n}{2n+1})$ and $\lim\limits_{n \to \infty} \ln ( \dfrac{n}{2n+1})=\ln \dfrac{1}{2} \ne 0$ Hence, the given series diverges as per the nth term integral test.