Convergence of $\sum_1^\infty \ln (\frac{3+n^p}{2+n^p})$ I am able to prove divergence for $p<0$ or $p=0$.
How can I prove convergence/divergence for $p>0$.

Let $x_n=\log((3+n^p)/(2+n^p))$. This is $x_n=\log(1+y_n)$ with $y_n=1/(2+n^p)$. 
If $p\lt0$, $y_n\to1/2$ and if $p=0$, $y_n\to1/3$. In both cases, $y_n\geqslant y^*$ for every $n$ with $y^*\gt0$ hence $x_n\geqslant\log(1+y^*)\gt0$ for every $n$ large enough and $\sum\limits_nx_n$ diverges. 
If $p\gt0$, $y_n\to0$ hence $y_n/2\leqslant\log(1+y_n)\leqslant y_n$ for every $n$ large enough. Since $y_n\leqslant1/n^p$ for every $n$ and $y_n\geqslant1/(2n^p)$ for every $n$ large enough, this shows that the series $\sum\limits_nx_n$ behaves like $\sum\limits_n1/n^p$ hence it diverges for every $0\lt p\leqslant1$ and it converges for every $p\gt1$.
Finally, the series $\sum\limits_nx_n=\sum\limits_n\log((3+n^p)/(2+n^p))$ converges if and only if $p\gt1$.
