Is The Series $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$ Divergent This is my solution for $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$. First I let the sequence in the series be labeled $A$, then I constructed a new sequence ($B$) that would have the similar behavior as $A$. This gives be the following:
$$A=\frac{n^4-3n+2}{4n^5+7}$$
$$B=\frac{1}{n}$$
Next I compare the value of the two sequances at $n=1$. This gives me $A=0$ and $B=1$, using this information I can concluded that $B \geq A$. Since I know that the $\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent, then so must $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$.
Does this solution make sense? I feel like i am missing something

So, not quite. Suppose you are considering two series $\sum a_n$ and $\sum b_n$ consisting of non-negative terms. If you know that $\sum a_n$ is divergent and it happens that $ a_n \leq b_n$ for every $n$ then  you may conclude that $\sum a_n \leq \sum b_n$. However, since you know the first series is divergent and consists of non-negative terms, you may conclude the same about the second series. But, this only works whenever the divergent series is the "smaller" one in the inequality. 
To answer your question, we would likely have to use a limit comparison test. 
Observe that $$\lim_{n\to \infty} \frac{\frac{n^4 - 3n +2}{4n^5 + 7}}{\frac{1}{n}} = \lim_{n\to \infty} \frac{n^5 -3n^2 +2n}{4n^5 + 7} = \lim_{n\to \infty} \frac{n^5 -3n^2 +2n}{4n^5 + 7}\cdot \frac{\frac{1}{n^5}}{\frac{1}{n^5}} = \lim_{n\to \infty} \frac{1 - \frac{3}{n^3} + \frac{2}{n^4}}{4 + \frac{7}{n^5}} = \frac{1}{4}$$
Since the limit is a positive, finite number we may conclude by the limit comparison test that since the harmonic series diverges, so must the series you were given. 
