1. **State the problem:** Determine the convergence or divergence of the series using the Comparison Test. 2. **Recall the Comparison Test:** - If $0 \leq a_n \leq b_n$ for all $n$ large enough and $\sum b_n$ converges, then $\sum a_n$ converges. - If $a_n \geq b_n \geq 0$ for all $n$ large enough and $\sum b_n$ diverges, then $\sum a_n$ diverges. 3. **Analyze each series:** - For $\sum \frac{1}{n \ln(n)}$: Given $\frac{1}{n \ln(n)} > \frac{2}{n}$ and $\sum \frac{2}{n}$ diverges (harmonic series), by the Comparison Test, $\sum \frac{1}{n \ln(n)}$ diverges. - For $\sum \frac{\ln(n)}{n}$: Given $\frac{\ln(n)}{n} > \frac{1}{n}$ and $\sum \frac{1}{n}$ diverges, by the Comparison Test, $\sum \frac{\ln(n)}{n}$ diverges. - For $\sum \frac{n}{(3-n)^2}$: Given $\frac{n}{(3-n)^2} < \frac{1}{n}$ and $\sum \frac{1}{n}$ diverges, but this is incorrect since $\sum \frac{1}{n}$ diverges, so the Comparison Test cannot conclude convergence here. The problem states it converges, so likely a typo or different comparison. - For $\sum \frac{1}{n^2}$: Given $\frac{1}{n^2} < \frac{1}{n^2}$ is false (equal), but $\sum \frac{1}{n^2}$ converges (p-series with $p=2>1$), so $\sum \frac{1}{n^2}$ converges. - For $\sum \frac{\ln(n)}{n^2}$: Given $\frac{\ln(n)}{n^2} > \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, by the Comparison Test, $\sum \frac{\ln(n)}{n^2}$ converges. - For $\sum \frac{\arctan(n)}{n^2}$: Given $\frac{\arctan(n)}{n^2} < \frac{\pi/2}{n^2}$ and $\sum \frac{1}{n^2}$ converges, by the Comparison Test, $\sum \frac{\arctan(n)}{n^2}$ converges. **Final answers:** - $\sum \frac{1}{n \ln(n)}$ diverges. - $\sum \frac{\ln(n)}{n}$ diverges. - $\sum \frac{n}{(3-n)^2}$ convergence is questionable based on given info. - $\sum \frac{1}{n^2}$ converges. - $\sum \frac{\ln(n)}{n^2}$ converges. - $\sum \frac{\arctan(n)}{n^2}$ converges.