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 Solution
Okay, 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 2
So, 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 3
So, 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 4
Now, 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 5
To 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 6
The interval of convergence is below and for summary purposes the radius of convergence is also shown.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\rm{Interval : }}\frac{9}{4} \le x \le \frac{{15}}{4}}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{3}{4}}}\]