Okay, let’s first see if the series converges or diverges if we put absolute value on the series terms.

\[\sum\limits_{n = 3}^\infty {\left| {\frac{{{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right)}}{{{n^3} + 1}}} \right|} = \sum\limits_{n = 3}^\infty {\frac{{n + 1}}{{{n^3} + 1}}} \]

We know by the \(p\)-series test that the following series converges.

\[\sum\limits_{n = 3}^\infty {\frac{1}{{{n^2}}}} \]

If we now compute the following limit,

\[c = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{n + 1}}{{{n^3} + 1}}\,\frac{{{n^2}}}{1}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^3} + {n^2}}}{{{n^3} + 1}}\,} \right] = 1\]

we know by the Limit Comparison Test that the two series in this problem have the same convergence because \(c\) is neither zero or infinity and because \(\sum\limits_{n = 3}^\infty {\frac{1}{{{n^2}}}} \) converges we know that the series from the problem statement must also converge.

Show Step 2