QUESTION $2$

The series ${\displaystyle \underset{n=2}{\overset{}{}}}(\frac{1}{n^{\frac{3}{2}}}+\frac{n^2}{n^3+1})$

A$.$ is divergent because ${\displaystyle \underset{n=2}{\overset{}{}}}\frac{1}{n^{\frac{3}{2}}}$ diverges by the p $-$series test and ${\displaystyle \underset{n=2}{\overset{}{}}}\frac{n^2}{n^3+1}$ converges by the integral test

B$.$ is convergent because ${\displaystyle \underset{n=2}{\overset{}{}}}\frac{1}{n^{\frac{3}{2}}}$ converges by p $-$series test and ${\displaystyle \underset{n=2}{\overset{}{}}}\frac{n^2}{n^3+1}$ converges by the integral test

C$.$ converges necessarily to the same value of the integral ${\displaystyle \underset{2}{\overset{}{}}}(\frac{1}{x^{\frac{3}{2}}}+\frac{x^2}{x^3+1})dx$

D$.$ is divergent because ${\displaystyle \underset{n=2}{\overset{}{}}}\frac{1}{n^{\frac{3}{2}}}$ converges by p $-$series test and ${\displaystyle \underset{n=2}{\overset{}{}}}\frac{n^2}{n^3+1}$ diverges by the integral test

E$.$ is divergent because ${\displaystyle \underset{n=2}{\overset{}{}}}\frac{1}{n^{\frac{3}{2}}}$ diverges by p $-$series test and ${\displaystyle \underset{n=2}{\overset{}{}}}\frac{n^2}{n^3+1}$ diverges by the integral test