convergence with comparison test $$\sum_{n=1}^\infty \frac{4n}{n^4+2n+9}$$
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!
 A: 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.
A: First we try to bound the series above:
\begin{align}
\sum_{n=1}^N\frac{1}{n^2}
&=\sum_{n=2}^N\frac{1}{n^2}\\
&<1+\sum_{n=2}^N\frac{1}{n(n-1)} \\
&=1+\sum_{n=2}^N(\frac{1}{n-1}-\frac{1}{n})\\
&=1+1+\frac{1}{N}\overset{N\to\infty}{=}2
\end{align}
By comparison test, since $$\sum_{n=1}^\infty\frac{1}{n^3}\le\sum_{n=1}^\infty\frac{1}{n^2}\le2$$
The following series convers:
$$\sum _{n=1}^{\infty}\frac{1}{n^3}\tag*{$\text{method (1)}$}$$

By p-series test, since $p=3>1$, the following series converges:
$$\sum _{n=1}^{\infty}\frac{1}{n^3}\tag*{$\text{method (2)}$}$$

By integral test, since $\int_1^{\infty}\frac{1}{x^3}dx=\frac{1}{2}$ converges, imples the following series conveges:
$$\sum _{n=1}^{\infty \:}\frac{1}{n^3}\tag*{$\text{method (3)}$}$$

Since $n^4+2n+9>n^4$, then we have
$$\sum _{n=1}^{\infty \:}\frac{n}{n^4+2n+9}\le \sum _{n=1}^{\infty \:}\frac{n}{n^4}=\sum _{n=1}^{\infty \:}\frac{1}{n^3}$$
By comparison test, the following series convers:
$$\sum _{n=1}^{\infty \:}\frac{n}{n^4+2n+9}=L\text{, where } L\in\mathbb{R}$$
Which implies:
$$\sum _{n=1}^{\infty \:}\frac{4n}{n^4+2n+9}=4L\in\mathbb{R}$$
Therefore $\sum _{n=1}^{\infty \:}\frac{4n}{n^4+2n+9}$ converges
