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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.

          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.

Question: Which of the following is the correct conclusion for the conver-/divergence of ∑n=4^∞(n-1)/(n^3+2) using the Comparison Test:
A: Because (n-1)/(n^3+2) > (1)/(n^2) and ∑n=4^∞(1)/(n^2) converges by p-series, ∑n=4^∞(n-1)/(n^3+2) also converges by the Comparison Test.
B: Because (n-1)/(n^3+2) < (1)/(n^2) and ∑n=4^∞(1)/(n^2) converges by p-series, ∑n=4^∞(n-1)/(n^3+2) also converges by the Comparison Test.
C: Because (n-1)/(n^3+2) < (1)/(n^2) and ∑n=4^∞(1)/(n^2) diverges by p-series, ∑n=4^∞(n-1)/(n^3+2) also diverges by the Comparison Test.
D: Because (n-1)/(n^3+2) > (1)/(n^2) and ∑n=4^∞(1)/(n^2) diverges by p-series, ∑n=4^∞(n-1)/(n^3+2) also diverges by the Comparison Test.
E: Because (n-1)/(n^3+2) = (1)/(n^2) and ∑n=4^∞(1)/(n^2) converges by p-series, ∑n=4^∞(n-1)/(n^3+2) also converges by the Comparison Test.

Added by Eric L.

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