Series, divergent $\sum_{n=1}^\infty\frac{n^4-3n}{n^5+n}$
I want to show, that this series is divergent, but i can not find a divergent minorant...
I tried 
$\sum_{n=1}^\infty\frac{n^4-3n}{n^5+n}=\sum_{n=1}^\infty\frac{n^3-3}{n^4+1}\geq\sum_{n=1}^\infty\frac{n^3}{n^4}=\sum_{n=1}^\infty\frac{1}{n}$
But that is wrong, i think. I have to find a minorant and not a majorant.
If someone could help me, that would be great.
Thanks.
 A: For $n\ge 2$, we have $n^4-3n\ge \frac{n^4}{2}$. 
Also, $n^5+n\le 2n^5$.
Thus for $n\ge 2$, our function is $\ge \frac{1}{4n}$. 
A: Alternately, You can use the
Limit Comparison Test
$$\sum_{n=n_0}^{\infty} a_n, \sum_{n=n_0}^{\infty} b_n, \lim_{n\to\infty} \frac{a_n}{b_n} \ne 0$$ 
With $a_n$ and $b_n$ series with positive terms; 
If $\sum_{n=n_0}^{\infty} b_n$ converges/diverges, $\sum_{n=n_0}^{\infty} a_n$ converges/diverges respectively. 

$$a_n = \frac{n^4 - 3n}{n^5 + n} = \frac{n^4}{n^5} \cdot \frac{1 - 3n/n^4}{1 + n/n^5} \approx \frac{1}{n}$$
($\frac{3n}{n^4}$ and $\frac{n}{n^5}$ $\to 0$ for $n$ very, very large)
Take $b_n = \frac{1}{n}$
$$\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{n^4 - 3n}{n^5 + n} \cdot n = \lim_{n\to\infty} \frac{n^5 - 3n^2}{n^5 + n} = \lim_{n\to\infty} \frac{n^5/n^5 - 3n/n^5}{n^5/n^5 + n/n^5} = \frac{1+0}{1+0} = 1 \ne 0$$
$\sum \frac{1}{n}$ diverges since its a p-series with $p = 1$.
Since all the criteria is met and $\sum b_n (\frac{1}{n})$ diverges, $\sum \frac{n^4 - 3n}{n^5 + n}$ diverges.
