How to prove if the series $\sum\limits_{n=2}^{\infty} \frac{n+1}{n^3-1} $ converges or diverges with comparison test?

Question

I have series $\sum_{n=2}^{\infty} \frac{n+1}{n^3-1} $

I think that comparison test would give the easiest solution but i'm not sure how to apply it. I know that i need to find larger sum to prove convergence or smaller to prove divergence, so

$\frac{1}{n} $ would give us larger sum (correct me if i am wrong), and we know that $\sum_{n=0}^{\infty} \frac{1}{n} $ diverges, therefore our series from beginning of question diverges, but wolfram alpha said that $\sum_{n=2}^{\infty} \frac{n+1}{n^3-1} $ by comparison test converges.

Where am i making mistake, and what is the best way to apply comparison test on $\sum_{n=2}^{\infty} \frac{n+1}{n^3-1} $?

Answer 1

Compare the given series with $\sum_{n=1}^\infty v_n=\sum\dfrac{1}{n^2}$.

If $u_n=\dfrac{n+1}{n^3+1}$ and $v_n=\dfrac{1}{n^2}$

then $\dfrac{u_n}{v_n}=\dfrac{(n^2)( n+1)}{n^3+1}=\dfrac{1+\dfrac{1}{n}}{1+\dfrac{1}{n^3}}\to 1<\infty$ as $n\to \infty$

Since $\sum_{n=1}^\infty \dfrac{1}{n^2}$ converges so does $\sum u_n=\sum\dfrac{n+1}{n^3+1}$

Answer 2

What about exploiting creative telescoping? For any $n\geq 2$ we have $$ 0\leq \frac{n+1}{n^3-1} \leq \frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}\tag{1}$$ hence: $$ 0\leq \sum_{n=2}^{N}\frac{n+1}{n^3-1}\leq \sum_{n=2}^{N}\frac{1}{n(n-1)} = 1-\frac{1}{N}<1.\tag{2}$$