proving convergence and calculating sum of a series So I have this series:
$$\displaystyle\sum_{n=1}^{\infty}\frac{3n^2+15n+9}{n^4+6n^3+9n^2}$$
I noticed:
$$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}(\frac{3n^2+15n+9}{n^2+6n+9})$$
So can I say that it's convergent already or should I use a critera on the second fraction in the sum?
Also for sum I could use partial fractions or am I mistaken?
Any help would be appreciated.

Consider 
\begin{align}
a_n=\frac{3n^2+15n+9}{n^4+6n^3+9n^2} &\le \frac{3n^2+15n^2+9n^2}{n^2(n+3)^2} \\
&= \frac{27}{(n+3)^2}\\
&\le \frac{27}{n^2} = b_n
\end{align}
It is well know that
$$\sum_{n=1}^\infty \frac{1}{n^2}$$
converges. Since $0\le a_n\le b_n$ and $\sum b_n$ converges, we may conclude that $\sum a_n$ converges.
