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 10.10 : Ratio Test
5. Determine if the following series converges or diverges.
\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{6n + 7}}} \]Show All Steps Hide All Steps
Start SolutionWe’ll need to compute \(L\).
\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{a_{n + 1}}\frac{1}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 1 + 1}}}}{{6\left( {n + 1} \right) + 7}}\frac{{6n + 7}}{{{{\left( { - 1} \right)}^{n + 1}}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 2}}}}{{6n + 13}}\frac{{6n + 7}}{{{{\left( { - 1} \right)}^{n + 1}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( { - 1} \right)\left( {6n + 7} \right)}}{{6n + 13}}} \right| = 1\end{align*}\]When computing \({a_{n + 1}}\) be careful to pay attention to any coefficients of \(n\) and powers of \(n\). Failure to properly deal with these is one of the biggest mistakes that students make in this computation and mistakes at that level often lead to the wrong answer!
Show Step 2Okay, we can see that \(L = 1\) and so by the Ratio Test tells us nothing about this series.
Show Step 3Just because the Ratio Test doesn’t tell us anything doesn’t mean we can’t determine if this series converges or diverges.
In fact, it’s actually quite simple to do in this case. This is an Alternating Series with,
\[{b_n} = \frac{1}{{6n + 7}}\]The \({b_n}\) are clearly positive and it should be pretty obvious (hopefully) that they also form a decreasing sequence. Finally, we also can see that \(\mathop {\lim }\limits_{n \to \infty } {b_n} = 0\) and so by the Alternating Series Test this series will converge.
Note, that if this series were not in this section doing this as an Alternating Series from the start would probably have been the best way of approaching this problem.