I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 10.9 : Absolute Convergence
3. Determine if the following series is absolutely convergent, conditionally convergent or divergent.
\[\sum\limits_{n = 3}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right)}}{{{n^3} + 1}}} \]Show All Steps Hide All Steps
Start SolutionOkay, 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 2So, because the series with the absolute value converges we know that the series in the problem statement is absolutely convergent.