Section 10.14 : Power Series
1. For the following power series determine the interval and radius of convergence.
\[\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 3} \right)}^{2 + n}}\left( {{n^2} + 1} \right)}}{{\left( {4x - 12} \right)}^n}} \]Show All Steps Hide All Steps
Start SolutionOkay, let’s start off with the Ratio Test to get our hands on \(L\).
\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( {4x - 12} \right)}^{n + 1}}}}{{{{\left( { - 3} \right)}^{3 + n}}\left( {{{\left( {n + 1} \right)}^2} + 1} \right)}}\frac{{{{\left( { - 3} \right)}^{2 + n}}\left( {{n^2} + 1} \right)}}{{{{\left( {4x - 12} \right)}^n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {4x - 12} \right)}}{{\left( { - 3} \right)\left( {{{\left( {n + 1} \right)}^2} + 1} \right)}}\frac{{\left( {{n^2} + 1} \right)}}{1}} \right|\\ & = \left| {4x - 12} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {{n^2} + 1} \right)}}{{3\left( {{{\left( {n + 1} \right)}^2} + 1} \right)}} = \frac{1}{3}\left| {4x - 12} \right|\end{align*}\] Show Step 2So, we know that the series will converge if,
\[\frac{1}{3}\left| {4x - 12} \right| < 1\hspace{0.25in} \to \hspace{0.25in}\frac{4}{3}\left| {x - 3} \right| < 1\hspace{0.25in} \to \hspace{0.25in}\left| {x - 3} \right| < \frac{3}{4}\] Show Step 3So, from the previous step we see that the radius of convergence is \(\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{3}{4}}}\).
Show Step 4Now, let’s start working on the interval of convergence. Let’s break up the inequality we got in Step 2.
\[ - \frac{3}{4} < x - 3 < \frac{3}{4}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in}\frac{9}{4} < x < \frac{{15}}{4}\] Show Step 5To finalize the interval of convergence we need to check the end points of the inequality from Step 4.
\(\displaystyle x = \frac{9}{4}\,\,\,:\,\,\,\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 3} \right)}^{2 + n}}\left( {{n^2} + 1} \right)}}{{\left( { - 3} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 3} \right)}^2}\left( {{n^2} + 1} \right)}}} = \sum\limits_{n = 0}^\infty {\frac{1}{{9\left( {{n^2} + 1} \right)}}} \)\(\displaystyle x = \frac{{15}}{4}\,\,\,:\,\,\,\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 1} \right)}^{2 + n}}{{\left( 3 \right)}^{2 + n}}\left( {{n^2} + 1} \right)}}{{\left( 3 \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 1} \right)}^{2 + n}}{{\left( 3 \right)}^2}\left( {{n^2} + 1} \right)}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^{2 + n}}}}{{9\left( {{n^2} + 1} \right)}}} \)
Now, we can do a quick Comparison Test on the first series to see that it converges and we can do a quick Alternating Series Test on the second series to see that is also converges.
We’ll leave it to you to verify both of these statements.
Show Step 6The interval of convergence is below and for summary purposes the radius of convergence is also shown.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\rm{Interval : }}\frac{9}{4} \le x \le \frac{{15}}{4}}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{3}{4}}}\]