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.
---------------
Let $a_{n}=\displaystyle \frac{n}{n^{2}+1}$.
...If the numerator had $n^2$ instead of $n$, the limit would be easily found, which is why we observe
$\displaystyle \sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}\frac{1}{n}$ , a divergent p-series
Test the hypothesis of the theorem:
$\displaystyle \lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}$ =$\displaystyle \lim_{n\rightarrow\infty}\frac{\frac{n}{n^{2}+1}}{\frac{1}{n}}=\lim_{n\rightarrow\infty}\frac{n^{2}}{n^{2}+1}$
... divide both numerator and denominator with $ n^{2}$...
$=\displaystyle \lim_{n\rightarrow\infty}\frac{1}{1+\frac{1}{n^{2}}}=1=L$ (finite and positive)
So,
$\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges as well (since they both diverge).