Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
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 \(\require{bbox} \bbox[2pt,border:1px solid black]{{R = 0}}\) .