Convergence of series $\sum\limits_{n=2}^\infty\frac{n^3+1}{n^4-1}$ 
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?

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