1. The problem involves determining the convergence or divergence of series using the Comparison Test. 2. The Comparison Test states: If $0 \leq a_n \leq b_n$ for all $n$ beyond some index, and $\sum b_n$ converges, then $\sum a_n$ converges. Conversely, if $\sum b_n$ diverges and $a_n \geq b_n \geq 0$, then $\sum a_n$ diverges. 3. For each series, we compare the given term $a_n$ with a simpler term $b_n$ whose convergence behavior is known. 4. Step-by-step analysis: 1. For $n>1$, $\frac{1}{n \ln(n)} < \frac{2}{n}$. Since $\sum \frac{2}{n}$ diverges (harmonic series), by Comparison Test, $\sum \frac{1}{n \ln(n)}$ diverges. 2. For $n>2$, $\frac{n}{n^3 - 9} < \frac{2}{n^2}$. Since $\sum \frac{2}{n^2}$ converges (p-series with $p=2>1$), $\sum \frac{n}{n^3 - 9}$ converges. 3. For $n>1$, $\frac{\arctan(n)}{n^3} < \frac{\pi/2}{n^3}$. Since $\sum \frac{\pi/2}{n^3}$ converges (p-series with $p=3>1$), $\sum \frac{\arctan(n)}{n^3}$ converges. 4. For $n>2$, $\frac{\ln(n)}{n} > \frac{1}{n}$. Since $\sum \frac{1}{n}$ diverges, $\sum \frac{\ln(n)}{n}$ diverges. 5. For $n>1$, $\frac{\ln(n)}{n^2} < \frac{1}{n^{1.5}}$. Since $\sum \frac{1}{n^{1.5}}$ converges, $\sum \frac{\ln(n)}{n^2}$ converges. 6. For $n>2$, $\frac{\ln(n)}{n^2} > \frac{1}{n^2}$. Since $\sum \frac{1}{n^2}$ converges, $\sum \frac{\ln(n)}{n^2}$ converges. 5. Summary: Using the Comparison Test and known series convergence/divergence, we conclude the behavior of each series as stated. Final answers: - $\sum \frac{1}{n \ln(n)}$ diverges. - $\sum \frac{n}{n^3 - 9}$ converges. - $\sum \frac{\arctan(n)}{n^3}$ converges. - $\sum \frac{\ln(n)}{n}$ diverges. - $\sum \frac{\ln(n)}{n^2}$ converges.