## Calculus (3rd Edition)

The series $\sum_{n=1}^{\infty} \frac{ 1}{\sqrt{n^2+1}}$ diverges.
Use the limit comparison test with the p-series $\sum_{n=1}^{\infty} \frac{1}{n}$ Since we have $$L=\lim_{n\to \infty } \frac{1/\sqrt{n^2+1}}{1/n }=\lim_{n\to \infty } \frac{ n }{\sqrt{n^2+1}}\\ =\lim_{n\to \infty } \frac{ 1}{\sqrt{1+1/n^2}}=1\gt0$$ then the series $\sum_{n=1}^{\infty} \frac{ 1}{\sqrt{n^2+1}}$ diverges because the p-series $\sum_{n=1}^{\infty} \frac{ 1}{ n }$, $p=1$, diverges.