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

Since, we have $\lim\limits_{n \to \infty} n \sin (\dfrac{1}{n})$ or, $\lim\limits_{n \to \infty} n \sin (\dfrac{1}{n})=\lim\limits_{n \to \infty} \dfrac{\sin (1/n)}{1/n})$ or, $=\lim\limits_{x \to 0} \dfrac{\sin x}{x}$ or, $=1 \ne 0$ Hence, the series is Divergent by the n-th term Test of Divergence