# 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.

• But the series indeed diverges. You can do the limit comparison test with the harmonic series. That's very easy – imranfat Nov 8 '13 at 2:45
• make this an answer. – ncmathsadist Nov 8 '13 at 2:45
• @imranfat Yeah, that's right. But how come the comparison test doesn't work? – Alti Nov 8 '13 at 2:46
• I am not saying that the comparison test would not work. But then you need to show that the given series is (ultimately) greater than the harmonic series for some k and beyond. That's not so easy, particularly because the denominator has a plus sign. When you have a given series consisting of a rational fraction, i.e. polynomial over polynomial, then the difference of their degrees is what you should compare it with. That tells you convergence/divergence. The limit comparison test will give you a finite limit L, hence the problem is solved. – imranfat Nov 8 '13 at 2:50

Hint: For $k\ge3$, $$\frac{k^2-1}{k^3+4}\ge\frac1{k+1}$$

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.

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

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