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?

IMHO this could be written more precisely. For example the second inequality does not look right. I would suggest the following.
$$\frac{3k^2+k+1}{k^4+k^3+4}=\frac{k^2}{k^4} \cdot \frac{3+1/k +1/k^2}{1+1/k} \frac{1}{k^2} \cdot \frac{3 + 1+ 1}{1+1}. $$
Now one can conclude that each of the entries of the sum is smaller or equal to $\frac{5}{2k^2}$. Therefore, according to your readings, the sum converges.
