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