Answer
See the explanation
Work Step by Step
Part (a)
Observe that
$\frac{n^2+n}{n^3-2}>\frac{n^2}{n^3-2}$
because the left side has a bigger nominator and observe that
$\frac{n^2}{n^3-2}>\frac{n^2}{n^3}$
because the left side has a smaller denominator.
Combining those conditions, we get
$\frac{n^2+n}{n^3-2}>\frac{n^2}{n^3}$
or
$\frac{n^2+n}{n^3-2}>\frac{1}{n}$
We know that $\sum_{n=2}^\infty \frac{1}{n}$ is a harmonic series and it is a divergent series.
Therefore, using the Direct Comparison Test wit $a_n=\frac{n^2+n}{n^3-2}$ and $b_n=\frac{1}{n}$ we conclude that the series $\sum_{n=2}^\infty \frac{n^2+n}{n^3-2}$ diverges.
Part (b)
We use the Limit Comparison Test with
$a_n=\frac{n^2+n}{n^3+2}$ and $b_n=\frac{1}{n}$
and obtain
$\lim\limits_{n \to \infty}\frac{a_n}{b_n}=\lim\limits_{n \to \infty}\frac{\frac{n^2+n}{n^3+2}}{\frac{1}{n}}=\lim\limits_{n \to \infty}\frac{n^3+n^2}{n^3+2}=\lim\limits_{n \to \infty}\frac{1+1/n}{1+2/n^3}=\frac{1+0}{1+0}=1>0$
Since this limit exists and $\sum_{n=2}^\infty \frac{1}{n}$ is a divergent series, it follows by the Limit Comparison Test that the series $\sum_{n=2}^\infty \frac{n^2+n}{n^3+2}$ diverges.
