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

Consider $\lim\limits_{n \to \infty} \dfrac{a_n}{b_n}=\lim\limits_{n \to \infty} \dfrac{1/n\sqrt {n^2+1}}{1/n^2}$ or, $\lim\limits_{n \to \infty} \dfrac{a_n}{b_n}=\lim\limits_{n \to \infty} \dfrac{n}{\sqrt {n^2+1}}$ or, $\lim\limits_{n \to \infty} \dfrac{a_n}{b_n}=\lim\limits_{n \to \infty} \dfrac{n/n}{\sqrt {n^2/n^2+1/n^2}}$ or, $\lim\limits_{n \to \infty} \dfrac{a_n}{b_n}=\dfrac{1}{\sqrt {(1+0)}}=1$ Thus, the series Converges Absolutely by the Limit Comparison Test.