Why is $\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1}$ divergent

Question

I am stuck on this problem. I have checked on wolfram alpha, and the calculator says that this series is divergent. However, I can't find a way to prove it.

The problem:

$$\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1}$$

So I tried doing this problem with the direct comparison test.

I have concluded that

$$
\frac{n^2}{n^3} >\frac{n^2-1}{n^3+1}
$$
$$
b_{n}> a_{n}
$$

$b_{n}$ is a p-series, where p=1. This means that the series $b_{n}$ is divergent, making the test direct comparison test inconclusive.
$$
\frac{n^2}{n^3} = \frac{1}{n}
$$

I appreciate any help… thanks! 🙂

Answer 1

Note that

$$\frac{n^2-1}{n^3+1} \geq \frac{n^2 - \frac{n^2}{2}}{n^3+n^3} = \frac{1}{4}\frac{n^2}{n^3} =\frac{1}{4n}. $$