Answer
(c)
Work Step by Step
To show that the series $\sum_{n=1}^\infty \frac{n}{n^2+1}$ diverges by the Direct Comparison Test, we consider $a_n=\frac{n}{n^2+1}$ and then we need to find a divergent series $\sum_{n=1}^\infty b_n$ satisfying $a_n\geq b_n$ for all $n\geq 1$.
It means that the inequality (b) can not be used and the inequality (a) can not be used since the series $\sum_{n=1}^\infty \frac{1}{n^2+1}$ is convergent.
Thus, the inequality that can be used is (c) $\frac{n}{n^2+1}\geq \frac{1}{2n}$ since the series $\sum_{n=1}^\infty \frac{1}{2n}=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n}$ is divergent (this series is a constant times the $p-$series with $p=1$).
