# Comparison test for $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}}$

I cannot seem to find a divergent series that is smaller than $\frac{1}{\sqrt{n^2 + 1}}$ to compare it with, while Wolfram Alpha does. How can I use the comparison test to prove that the series diverges?

• ${1\over\sqrt{n^2+1}}\ge{1\over\sqrt{n^2+n^2}} ={1\over\sqrt {2n^2}}$. – David Mitra Dec 22 '13 at 12:58

## 2 Answers

Hint: for $n\ge1$, $${1\over\sqrt{n^2+1}}\ge{1\over\sqrt{n^2+n^2}} ={1\over\sqrt {2n^2}}={1\over\sqrt2}\cdot{1\over n\vphantom{\sqrt2}}.$$

You can simply use the asymtotic comparison $$\frac{1}{\sqrt{n^2+1}}\sim_\infty \frac 1 n$$