<s>[INST] Why is $\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1}$ divergent? 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! :)
 [/INST] $$
\frac{n^2-1}{n^3+1} = \frac{n^2+1 -2}{n^3+1} = \frac{n^2+1}{n^3+1} -\frac{2}{n^3+1} \geq \frac{1}{n} - \frac{2}{n^3+1} \quad \forall\ n \in \mathbb{N}.
$$
Therefore,
\begin{align}\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1} = \sum\limits_{n=1}^{\infty}\left(\frac{n^2+1}{n^3+1} -\frac{2}{n^3+1}\right)\\
\\
= \sum\limits_{n=1}^{\infty}\frac{n^2+1}{n^3+1}\ -\ \sum\limits_{n=1}^{\infty}\frac{2}{n^3+1} \\
\\
\geq \left(\sum\limits_{n=1}^{\infty} \frac{1}{n}\right)\ -\ L \\
\end{align}
for some $L\in \mathbb{R}.$
</s>