Show with comparison test that this series converges: $\sum_{n=17}^{\infty}\frac{7n^{2}+18n+2}{n^{4}+100}$ 
Show with comparison test that this series converges:
  $\sum_{n=17}^{\infty}\frac{7n^{2}+18n+2}{n^{4}+100}$

$$\frac{7n^{2}+18n+2}{n^{4}+100}>\frac{7n^{2}+7n}{n^{4}}=\frac{7}{n^{2}}+\frac{7}{n^{3}}>\frac{7}{n^{2}}$$
$$\Rightarrow \sum_{n=17}^{\infty}\frac{7}{n^{2}}=7\sum_{n=17}^{\infty}\frac{1}{n^{2}}$$
We know that $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ converges, so the complete series will converge, too.
Did I do it right?

You need to find an upper bound, because having a lower bound that converges does not give you any information at all, since the series could diverge or converge. This is one way of doing things, there are many others:
$$7n^2+18n+2<7(n+4)^2<7(2n)^2=28n^2$$
In addition $n^4+100>n^4$
So $$\frac{7n^2+18n+2}{n^4+100}<\frac {28n^2}{n^4}$$
 $$\sum_{n=17}^\infty\frac{7n^2+18n+2}{n^4+100}<\sum_{n=17}^\infty\frac {28n^2}{n^4}=\sum_{n=17}^\infty\frac {28}{n^2}$$
Hence, it does converge.
