Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter I (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.)

1. For all $n > 1$, $\frac{\ln (n)}{n^2} < \frac{1}{n^{1.5}}$, and the series $\sum \frac{1}{n^{1.5}}$ converges, so by the Comparison Test, the series $\sum \frac{\ln (n)}{n^2}$ converges.
2. For all $n > 1$, $\frac{\arctan (n)}{n^3} < \frac{\pi}{2n^3}$, and the series $\frac{\pi}{2} \sum \frac{1}{n^3}$ converges, so by the Comparison Test, the series $\sum \frac{\arctan (n)}{n^3}$ converges.
3. For all $n > 2$, $\frac{1}{n^2- 5} < \frac{1}{n^2}$, and the series $\sum \frac{1}{n^2}$ converges, so by the Comparison Test, the series $\sum \frac{1}{n^2 - 5}$ converges.
4. For all $n > 2$, $\frac{\ln (n)}{n} > \frac{1}{n}$, and the series $\sum \frac{1}{n}$ diverges, so by the Comparison Test, the series $\sum \frac{\ln (n)}{n}$ diverges.
5. For all $n > 2$, $\frac{n}{n^3 - 2} < \frac{2}{n^2}$, and the series $2 \sum \frac{1}{n^2}$ converges, so by the Comparison Test, the series $\sum \frac{n}{n^3 - 2}$ converges.
6. For all $n > 1$, $\frac{n}{2-n^3} < \frac{1}{n^2}$, and the series $\sum \frac{1}{n^2}$ converges, so by the Comparison Test, the series $\sum \frac{n}{2 -n^3}$ converges.

In order to get credit for this problem all answers must be correct.