TITLE: convergence with comparison test
QUESTION [5 upvotes]: $$\sum_{n=1}^\infty \frac{4n}{n^4+2n+9}$$
Hi guys! I need to use comparison test on this series, I haven't done a lot of comparison tests so far, so I'm not sure what to compare it with. Should I use $$\frac{4n}{n^4}$$ which is $$\frac{4}{n^3},$$ but then I have to show convergence of $$\frac{4}{n^3}.$$ Do I keep comparing until I get to $$\frac{1}{n^2}$$ which I know and shown its convergence at a previous exercise?
If anyone can help me with it, I would greatly appreciate it. Thank you!

REPLY [2 votes]: More simply use limit comparison test with $\sum \frac1{n^2}$ and since
$$\frac{\frac{4n}{n^4+2n+9}}{\frac1{n^2}}= \frac{4n^3}{n^4+2n+9}\to0$$
we conclude that the series converges.