calculus - Convergence of series $sumlimits_{n=2}^inftyfrac{n^3+1}{n^4-1}$

Investigate the series for convergence and if possible, determine its
limit: $\sum\limits_{n=2}^\infty\frac{n^3+1}{n^4-1}$

My thoughts

Let there be the sequence $s_n = \frac{n^3+1}{n^4-1}, n \ge 2$.

I have tried different things with no avail. I suspect I must find a lower series which diverges, in order to prove that it diverges, and use the comparison test.

Could you give me some hints as a comment? Then I'll try to update my question, so you can double-check it afterwards.

Update

$$s_n \gt \frac{n^3}{n^4} = \frac1n$$

which means that

$$\lim\limits_{n\to\infty} s_n > \lim\limits_{n\to\infty}\frac1n$$

but $$\sum\limits_{n=2}^\infty\frac1n = \infty$$

so $$\sum\limits_{n=2}^\infty s_n = \infty$$

thus the series $\sum\limits_{n=2}^\infty s_n$ also diverges.

The question is: is this formally sufficient?

Answer

$$\frac{n^3+1}{n^4-1}\gt\frac{n^3}{n^4}=\frac1n\;.$$