Answer
Diverges
Work Step by Step
Th. 9.13 Limit Comparison Test:
If $a_{n} > 0, b_{n} > 0$, and $\displaystyle \lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=L,$
where $L$ is finite and positive, then
$\displaystyle \sum_{n=1}^{\infty}a_{n}$ and $\displaystyle \sum_{n=1}^{\infty}b_{n}$ either both converge or both diverge.
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Let $a_{n}=\displaystyle \frac{1}{\sqrt{n^{2}+1}}$.
...If the numerator had a factor $n$, (and, $ n=\sqrt{n^{2}}$ for positive $n$) the limit would be easily found, which is why we observe
$\displaystyle \sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}\frac{1}{n}$
which is a divergent p-series (p=1).
Test the hypothesis of the theorem:
$\displaystyle \lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}$ =$\displaystyle \lim_{n\rightarrow\infty}\frac{\frac{1}{\sqrt{n^{2}+1}}}{\frac{1}{n}}=\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^{2}+1}}$
... divide both numerator and denominator with $ n$ ...
$=\displaystyle \lim_{n\rightarrow\infty}\frac{1}{\sqrt{1+\frac{1}{n^{2}}}}=1=L$ (finite and positive)
So,
$\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges as well (since they both diverge).