Answer
See explanation
Work Step by Step
Part (a)
Observe that
$\frac{n}{n^3+5}<\frac{n}{n^3}$
because the left side has a bigger denominator.
Now, we have:
$\frac{n}{n^3+5}<\frac{1}{n^2}$ for all $n\geq 1$
Apply the Direct Comparison Test with $a_n=\frac{n}{n^3+5}$ and $b_n=\frac{1}{n^2}$.
Since $\sum_{n=2}^\infty \frac{1}{n^2}$ is convergent $p-$series with $p=2$, it follows by the Direct Comparison Test that the series $\sum_{n=2}^\infty \frac{n}{n^3+5}$ converges.
Part (b)
We use the Limit Comparison Test with
$a_n=\frac{n}{n^3-5}$ and $b_n=\frac{1}{n^2}$
and obtain
$\lim\limits_{n \to \infty}\frac{a_n}{b_n}=\lim\limits_{n \to \infty}\frac{\frac{n}{n^3-5}}{\frac{1}{n^2}}=\lim\limits_{n \to \infty}\frac{n^3}{n^3-5}=\lim\limits_{n \to \infty}\frac{1}{1-5/n^3}=\frac{1}{1-0}=1>0$.
Since this limit exists and $\sum_{n=2}^\infty \frac{1}{n^2}$ is a convergent $p-$series (with $p=2$), it follows by the Limit Comparison Test that the series $\sum_{n=2}^\infty\frac{n}{n^3-5}$ converges.
