Answer
Divergent
Work Step by Step
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 Limit Test.
Hence, the series is divergent
