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. 
 A: Hint: For $k\ge3$,
$$
\frac{k^2-1}{k^3+4}\ge\frac1{k+1}
$$
A: 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.
A: To show that the given series is (ultimately) greater than the harmonic series for some k and beyond, We could consider this limit:
$$\lim_{k\rightarrow\infty}\frac{\frac{1}{k}}{\frac{k^2-1}{k^3+4}} = \lim_{k\rightarrow\infty}\frac{k^3+4}{k^3-k}= 1$$
Hence, by the definition of limit,
$$\forall\varepsilon\gt0,\exists N\gt0,\text{such that} ~k \gt N,~~~~~\left|\frac{\frac{1}{k}}{\frac{k^2-1}{k^3+4}} - 1 \right| \lt \varepsilon$$
Let $\varepsilon = 1$, for $k \gt N$, we have
\begin{align*}
\ \left|\frac{\frac{1}{k}}{\frac{k^2-1}{k^3+4}} - 1\right| \lt 1\rightarrow 0 \lt \frac{\frac{1}{k}}{\frac{k^2-1}{k^3+4}} \lt 2 \rightarrow \frac{k^2-1}{k^3+4} \gt \frac{1}{2k}\rightarrow \sum_{k=N+1}^{\infty}\frac{k^2-1}{k^3+4} \gt \sum_{k=N+1}^{\infty}\frac{1}{2k} \space\space\space\text{(for k > N)}
\end{align*}
Since the series $\sum_{k=1}^{\infty}\frac{1}{2k}$ is divergent,by the comparison test, $ \sum_{k=1}^{\infty}\frac{k^2-1}{k^3+4}$ is divergent
A: If you use the limit comparison test instead, and compare it with the harmonic series, you will find for the limit L=1 ($k^3/k^3$), since the harmonic series diverges, so does your given series
