<s>[INST] $\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$ Divergent? Did I do this right? $$\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$$
$$a_n = \frac{1}{\sqrt{n^2+1}} <\frac{1}{\sqrt{n^2}} \le \frac{1}{n},  \forall n\ge1$$
Then by The Comparison Test with $b_n$, where $b_n$ is a divergent p-series  = $\frac{1}{n}$ and $a_n \le b_n$, 
Then
$\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$ is Divergent.  
 [/INST] This conclusion is not quite correct.  Note that $n^2+1>n^2 \implies \frac{1}{\sqrt{n^2+1}}<\frac1n$ is correct.  But $\frac{1}{n^2}\le \frac1n$ and $\sum_{n=1}^\infty \frac{1}{n^2}$ converges.
Instead, we can assert that for $n\ge 1$, $n^2+1\le 2n^2$.  Therefore, $\frac{1}{\sqrt{n^2+1}}>\frac{1}{\sqrt{2}\,n}$.  And since $\sum_{n=1}^\infty\frac1n$ diverges, so does $\sum_{n=1}^\infty\frac{1}{\sqrt{n^2+1}}$.
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