[tex]\displaystyle\sum_{n\ge0}\frac{\sin n}{n^3+1}[/tex]
You have
[tex]-\dfrac1{n^3+1}\le\dfrac{\sin n}{n^3+1}\le\dfrac1{n^3+1}[/tex]
with [tex]\pm\dfrac1{n^3+1}\to0[/tex] as [tex]n\to\infty[/tex], so by the squeeze theorem,
[tex]\displaystyle\lim_{n\to\infty}\frac{\sin n}{n^3+1}=0[/tex]
and so the [tex]n[/tex]th term test fails/doesn't confirm that the series diverges.
It does, however, converge by the comparison test. Consider the convergent
[tex]\displaystyle\sum_{n\ge1}\frac1{n^3}[/tex]
Because this converges, the alternating series
[tex]\displaystyle\sum_{n\ge1}\frac{(-1)^n}{n^3}[/tex]
converges absolutely, and
[tex]\left|\dfrac{\sin n}{n^3+1}\right|\le\dfrac1{n^3+1}<\dfrac1{n^3}[/tex]
