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

Consider $a_n= \tan (\dfrac{1}{n})$ and $b_n=\dfrac{1}{ n}$ Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{ \tan (\dfrac{1}{n})}{\dfrac{1}{ n}}$ Thus, we have to apply L-Hospital's rule. Then $=\lim\limits_{n \to \infty} \dfrac{\sec^2 (1/n)}{1}$ or, $\sec (0)=1 \ne 0 \ne \infty$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n}$ is a divergent series due to the p-series test with $p \leq 1$ Thus the series diverges by the limit comparison test.