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
1. Determine if the following series is absolutely convergent, conditionally convergent or divergent.
\[\sum\limits_{n = 2}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{n^3} + 1}}} \]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.
\[\sum\limits_{n = 2}^\infty {\left| {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{n^3} + 1}}} \right|} = \sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}} \]Now, notice that,
\[\frac{1}{{{n^3} + 1}} < \frac{1}{{{n^3}}}\]and we know by the \(p\)-series test that
\[\sum\limits_{n = 2}^\infty {\frac{1}{{{n^3}}}} \]converges.
Therefore, by the Comparison Test we know that the series from the problem statement,
\[\sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}} \]will 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.