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

Consider $a_n=\sqrt {\dfrac{ n+1}{n^2+2}}$ and $b_n=\dfrac{ 1}{\sqrt n}$ Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\sqrt {\dfrac{ n+1}{n^2+2}}}{\dfrac{ 1}{\sqrt n}}$ Thus, we have $=\lim\limits_{n \to \infty} \dfrac{\sqrt {n^2+n}}{\sqrt {n^2+2}}$ or, $=1$ Hence, the series Diverges (as $b_n=\dfrac{ 1}{\sqrt n}$ is a divergent series) due to the limit comparison test.