proving convergence and calculating sum of a series

You can conclude that the series converges by the comparison test. We can also calculate the sum. Note that $$\frac{3n^{2}+15n+9}{n^{4}+6n^{3}+9n^{2}}=\frac{1}{n^{2}}+\frac{1}{n}-\frac{1}{n+3}-\frac{1}{\left(n+3\right)^{2}} $$ hence $$\sum_{n\geq1}\frac{3n^{2}+15n+9}{n^{4}+6n^{3}+9n^{2}}=\sum_{n\geq1}\left(\frac{1}{n^{2}}+\frac{1}{n}-\frac{1}{n+3}-\frac{1}{\left(n+3\right)^{2}}\right)$$ $$=2+\frac{1}{2}+\frac{1}{4}+\frac{1}{3}+\frac{1}{9} =\frac{115}{36} $$ because the series telescopes.

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.