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 $p>0$.
 A: 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 $p\gt1$.
A: You got:
$$\ln \left( \frac{3+n^p}{2+n^p}\right) = \ln \left( 1+ \frac{1}{2+n^p}\right)$$
hence the well known asymptotics $\ln (1+y)\approx y$ as $y\to 0$ and $p>0$ yield:
$$\ln \left( \frac{3+n^p}{2+n^p}\right) \approx \frac{1}{2+n^p} \approx \frac{1}{n^p}\; ;$$
therefore your series converges if and only if $p>1$.
A: Hints:
$$ \log(1+x)=x+\mathcal{O}(x^2); \qquad \frac{3+n^p}{2+n^p}=1+\frac{1}{2+n^p} $$
Compare term-for-term with the usual $p$-series, $\sum n^{-p}$.
