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
2. Determine if the following series is absolutely convergent, conditionally convergent or divergent.
\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 3}}}}{{\sqrt n }}} \]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.
\[\sum\limits_{n = 1}^\infty {\left| {\frac{{{{\left( { - 1} \right)}^{n - 3}}}}{{\sqrt n }}} \right|} = \sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt n }}} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^{\frac{1}{2}}}}}} \]Now, by the by the \(p\)-series test we can see that this series will diverge.
Show Step 2So, at this point we know that the series in the problem statement is not absolutely convergent so all we need to do is check to see if it’s conditionally convergent or divergent. To do this all we need to do is check the convergence of the series in the problem statement.
The series in the problem statement is an alternating series with,
\[{b_n} = \frac{1}{{\sqrt n }}\]Clearly the \({b_n}\) are positive so we can use the Alternating Series Test on this series. It is hopefully clear that the \({b_n}\) are a decreasing sequence and \(\mathop {\lim }\limits_{n \to \infty } {b_n} = 0\).
Therefore, by the Alternating Series Test the series from the problem statement is convergent.
Show Step 3So, because the series with the absolute value diverges and the series from the problem statement converges we know that the series in the problem statement is conditionally convergent.