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

For sufficiently large $N_0$, we have  $n \ge N_0$, then $$n^4-3n+2\ge \frac12 n^4$$
and $$4n^5+7 \le 5n^5.$$
That is $n \ge N_0$, we have
$$\frac{n^4-3n+2}{4n^5+7}\ge \frac{\frac12n^4}{5n^5}=\frac1{10n}$$
$\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}= \sum_{n=1}^{N_0-1} \frac{n^4-3n+2}{4n^5+7} + \sum_{n=N_0}^{\infty} \frac{n^4-3n+2}{4n^5+7}\ge \sum_{n=1}^{N_0-1} \frac{n^4-3n+2}{4n^5+7} + \frac1{10}\sum_{n=N_0}^{\infty} \frac{1}{n}$
Hence the series diverges.
