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 10.14 : Power Series
2. For the following power series determine the interval and radius of convergence.
\[\sum\limits_{n = 0}^\infty {\frac{{{n^{2n + 1}}}}{{{4^{3n}}}}{{\left( {2x + 17} \right)}^n}} \]Show All Steps Hide All Steps
Start SolutionOkay, let’s start off with the Root Test to get our hands on \(L\).
\[L = \mathop {\lim }\limits_{n \to \infty } {\left| {\frac{{{n^{2n + 1}}}}{{{4^{3n}}}}{{\left( {2x + 17} \right)}^n}} \right|^{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^{2 + \frac{1}{n}}}}}{{{4^3}}}\left( {2x + 17} \right)} \right| = \left| {2x + 17} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{{n^{2 + \frac{1}{n}}}}}{{{4^3}}}\]Okay, we can see that , in this case, \(L\) will be infinite provided \(x \ne - \frac{{17}}{2}\) and so the series will diverge for \(x \ne - \frac{{17}}{2}\). We also know that the power series will converge for \(x = -\frac{{17}}{2}\) (this is the value of \(a\) for this series!).
Show Step 2Therefore, we know that the interval of convergence is \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{{17}}{2}}}\) and the radius of convergence is \(\require{bbox} \bbox[2pt,border:1px solid black]{{R = 0}}\) .