## Calculus: Early Transcendentals 8th Edition

Use Limit Comparison Test with $a_{n}=\frac{1}{\sqrt {n^{2}+1}}$ and $b_{n}=\frac{1}{n}$ $\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{\frac{1}{\sqrt {n^{2}+1}}}{\frac{1}{n}}$ $=\lim\limits_{n \to \infty}\frac{1}{\frac{{\sqrt {n^{2}+1}}}{n}}$ $=\lim\limits_{n \to \infty}\frac{1}{\frac{{\sqrt {n^{2}+1}}}{\sqrt n^{2}}}$ $=\lim\limits_{n \to \infty}\frac{1}{{\sqrt {1+1/n^{2}}}}$ $=\frac{1}{\sqrt {1+0}}$ $=1\ne 0 \ne \infty$ The given series diverges by the Limit Test.