Question: Which of the following is the correct conclusion for the conver-\gence/divergence of $\sum_{n=4}^{\infty} \frac{n-1}{n^3+2}$ using the Comparison Test:
A: Because $\frac{n-1}{n^3+2} > \frac{1}{n^2}$ and $\sum_{n=4}^{\infty} \frac{1}{n^2}$ converges by p-series, $\sum_{n=4}^{\infty} \frac{n-1}{n^3+2}$ also converges by the Comparison Test.
B: Because $\frac{n-1}{n^3+2} < \frac{1}{n^2}$ and $\sum_{n=4}^{\infty} \frac{1}{n^2}$ converges by p-series, $\sum_{n=4}^{\infty} \frac{n-1}{n^3+2}$ also converges by the Comparison Test.
C: Because $\frac{n-1}{n^3+2} < \frac{1}{n^2}$ and $\sum_{n=4}^{\infty} \frac{1}{n^2}$ diverges by p-series, $\sum_{n=4}^{\infty} \frac{n-1}{n^3+2}$ also diverges by the Comparison Test.
D: Because $\frac{n-1}{n^3+2} > \frac{1}{n^2}$ and $\sum_{n=4}^{\infty} \frac{1}{n^2}$ diverges by p-series, $\sum_{n=4}^{\infty} \frac{n-1}{n^3+2}$ also diverges by the Comparison Test.
E: Because $\frac{n-1}{n^3+2} = \frac{1}{n^2}$ and $\sum_{n=4}^{\infty} \frac{1}{n^2}$ converges by p-series, $\sum_{n=4}^{\infty} \frac{n-1}{n^3+2}$ also converges by the Comparison Test.
