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

Consider $a_n=\dfrac{1}{n+\sqrt n}$ and $b_n=\dfrac{1}{n}$ Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{1}{n+\sqrt n}}{\dfrac{1}{n}}$ Thus, we have $=\lim\limits_{n \to \infty} \dfrac{n}{n+\sqrt n}$ or, $=1$ Hence, the series diverges due to the limit comparison test.