How to solve this rational function with logarithm infinite series problem? 
Determine whether the series converges absolutely, conditionally, or diverges:
$\sum_{n=5}^\infty \frac{n^3+2n^2+1}{n^5-n^2+\ln n} $

I first thought of using the limit comparison test with a $\frac{1}{n^2}$ series and a $\frac{n^3+2n^2+1}{n^5-n^2+n} $ series. I found both converge. However, I realized that because $\ln n$ is always less than $n$, $\frac{n^3+2n^2+1}{n^5-n^2+\ln n} $ is always bigger than $\frac{n^3+2n^2+1}{n^5-n^2+n} $ so I can't use a series comparison test to finally solve this problem. What should I do to solve this problem?

Limit comparison test is not necessary here. A simple comparison test will suffice. Since $\ln n>0$ and for every $n>2$, the following holds for $n>5$:
$$\dfrac{n^3+2n^2+1}{n^5-n^2+\ln n}<\dfrac{n^3+2n^2+1}{n^5-n^2}<\dfrac{n^3+2n^2+n}{n^5-n^2}=\dfrac{2n^2+2n+2}{n(n-1)(n^2+n+1)}=\dfrac{2}{n(n-1)}$$
Then you can show that $\displaystyle\sum_{n=5}^{\infty}\dfrac{2}{n(n-1)}$ converges by the definition of infinite series, and thus the original series converges by comparison test.
