Test for convergence/divergence of $\sum_{k=1}^\infty \frac{k^2-1}{k^3+4}$ Given the series 

$$\sum_{k=1}^\infty \frac{k^2-1}{k^3+4}.$$ 

I need to test for convergence/divergence. I can compare this to the series $\sum_{k=1}^\infty \frac{1}{k}$, which diverges. 
To use the comparison test, won't I need to show that $\frac{k^2-1}{k^3+4}>\frac{k^3}{k^4}=\frac{1}{k}$, in order to state the original series diverges?
This doesn't seem to hold, I feel like I'm missing the obvious. 
Any help is appreciated. 
Thanks. 

When $k \geq 1$, we have $k^3+4 \leq k^3+4k^3 = 5k^3$ (Since $4 \leq 4k^3$ for $k \geq 1$). Also, when $k \geq 2$, we have $k^2-1 > \frac{k^2}{2}$ (This is again straightforward to verify, equality happens when $k = \sqrt{2}$ :)). Therefore,
$$\sum_{k=1}^\infty \frac{k^2-1}{k^3+4} = \sum_{k=2}^\infty \frac{k^2-1}{k^3+4} \geq \sum_{k=2}^\infty\frac{\frac{k^2}{2}}{5k^3} = \frac{1}{10}\sum_{k=2}^\infty\frac{1}{k}.$$
The lower bound is the harmonic series that clearly diverges. Hence the original series diverges.
