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
4. For the following power series determine the interval and radius of convergence.
\[\sum\limits_{n = 0}^\infty {\frac{{{4^{1 + 2n}}}}{{{5^{n + 1}}}}{{\left( {x + 3} \right)}^n}} \]Show All Steps Hide All Steps
Start SolutionOkay, let’s start off with the Ratio Test to get our hands on \(L\).
\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{4^{3 + 2n}}{{\left( {x + 3} \right)}^{n + 1}}}}{{{5^{n + 2}}}}\frac{{{5^{n + 1}}}}{{{4^{1 + 2n}}{{\left( {x + 3} \right)}^n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{4^2}\left( {x + 3} \right)}}{5}} \right| = \left| {x + 3} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{16}}{5} = \frac{{16}}{5}\left| {x + 3} \right|\] Show Step 2So, we know that the series will converge if,
\[\frac{{16}}{5}\left| {x + 3} \right| < 1\hspace{0.25in} \to \hspace{0.25in}\left| {x + 3} \right| < \frac{5}{{16}}\] Show Step 3So, from the previous step we see that the radius of convergence is \(\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{5}{{16}}}}\).
Show Step 4Now, let’s start working on the interval of convergence. Let’s break up the inequality we got in Step 2.
\[ - \frac{5}{{16}} < x + 3 < \frac{5}{{16}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in} - \frac{{53}}{{16}} < x < - \frac{{43}}{{16}}\] Show Step 5To finalize the interval of convergence we need to check the end points of the inequality from Step 4.
\( \displaystyle x = - \frac{{53}}{{16}}\,\,\,:\,\,\,\sum\limits_{n = 0}^\infty {\frac{{{4^1}{4^{2n}}}}{{{5^n}{5^1}}}{{\left( { - \frac{5}{{16}}} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{{4\left( {{{16}^n}} \right)}}{{{5^n}\left( 5 \right)}}\frac{{{{\left( { - 1} \right)}^n}{5^n}}}{{{{16}^n}}}} = \sum\limits_{n = 0}^\infty {\frac{{4{{\left( { - 1} \right)}^n}}}{5}} \)\(\displaystyle x = - \frac{{43}}{{16}}\,\,\,:\,\,\,\sum\limits_{n = 0}^\infty {\frac{{{4^1}{4^{2n}}}}{{{5^n}{5^1}}}{{\left( {\frac{5}{{16}}} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{{4\left( {{{16}^n}} \right)}}{{{5^n}\left( 5 \right)}}\frac{{{5^n}}}{{{{16}^n}}}} = \sum\limits_{n = 0}^\infty {\frac{4}{5}} \)
Now,
\[\mathop {\lim }\limits_{n \to \infty } \frac{{4{{\left( { - 1} \right)}^n}}}{5}\,\,\,\,\, - \,\,\,\,{\mbox{Does not exist}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{n \to \infty } \frac{4}{5} = \frac{4}{5}\]Therefore, each of these two series diverge by the Divergence Test.
Show Step 6The interval of convergence is below and for summary purposes the radius of convergence is also shown.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\rm{Interval : }} - \frac{{53}}{{16}} < x < - \frac{{43}}{{16}}}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{5}{{16}}}}\]