## Thomas' Calculus 13th Edition

Let $l=\lim\limits_{n \to \infty} \dfrac{p_n}{q_n}=\lim\limits_{n \to \infty} \dfrac{\dfrac{1}{n\sqrt {n^2+1}}}{\dfrac{1}{n^2}}$ and $\lim\limits_{n \to \infty} \dfrac{n}{\sqrt {n^2+1}}=\lim\limits_{n \to \infty} \dfrac{\dfrac{n}{n}}{\sqrt {\dfrac{n^2}{n^2}+\dfrac{1}{n^2}}}$ Thus, $l=\dfrac{1}{\sqrt {(1+0)}}=1$ Thus, the series Converges Absolutely by the Limit Comparison Test.