Decide if the series converges and prove it using comparison test: $\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}$ 
Decide if the series converges and prove it using comparison test:
  $\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}$

$$\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}< \frac{k^{2}+k}{k^{4}+k^{3}} < \frac{k^{2}}{k^{4}} \leq \frac{1}{k^{2}}$$
We know (from our readings) that $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$ is a converging series.
Thus the complete series will converge.
Did I do everything correctly?

Following your idea I would write
$$\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}< \frac{3k^{2}+3k}{k^{4}+k^{3}} = \frac{3k}{k^{3}} = \frac{3}{k^{2}}.$$
As regards your inequalities, how do you justify that 
$$\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}< \frac{k^{2}+k}{k^{4}+k^{3}}?$$
Notice that if $a$, $b$, $c$, $d$ are positive numbers
$$a<c\quad \mbox{and}\quad d<b\Rightarrow \frac{a}{b}<\frac{c}{d}.$$
