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

Consider $u_n=\dfrac{1}{\ln n}$ Here, $u_n$ is positive for all the values of $n$. Here, $\ln n$ is an increasing function, but the sequence $u_n$ is decreasing. Thus, $\lim\limits_{n \to \infty} u_n=\lim\limits_{n \to \infty} \dfrac{1}{\ln n}=0$ Hence, by the Alternating Series Test, the series converges.