Section 10.14 : Power Series
5. For the following power series determine the interval and radius of convergence.
\[\sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,{{\left( {4x - 1} \right)}^{n - 1}}} \]Show All Steps Hide All Steps
Start SolutionOkay, let’s start off with the Ratio Test to get our hands on \(L\).
\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{6^{n + 1}}{{\left( {4x - 1} \right)}^n}}}{{n + 1}}\,\frac{n}{{{6^n}{{\left( {4x - 1} \right)}^{n - 1}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{6n\left( {4x - 1} \right)}}{{n + 1}}} \right| = \left| {4x - 1} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{6n}}{{n + 1}} = 6\left| {4x - 1} \right|\] Show Step 2So, we know that the series will converge if,
\[6\left| {4x - 1} \right| < 1\hspace{0.25in} \to \hspace{0.25in}24\left| {x - \frac{1}{4}} \right| < 1\hspace{0.25in} \to \hspace{0.25in}\left| {x - \frac{1}{4}} \right| < \frac{1}{{24}}\] Show Step 3So, from the previous step we see that the radius of convergence is \(\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{1}{{24}}}}\).
Show Step 4Now, let’s start working on the interval of convergence. Let’s break up the inequality we got in Step 2.
\[ - \frac{1}{{24}} < x - \frac{1}{4} < \frac{1}{{24}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in}\frac{5}{{24}} < x < \frac{7}{{24}}\] Show Step 5To finalize the interval of convergence we need to check the end points of the inequality from Step 4.
\(\displaystyle x = \frac{5}{{24}}\,\,\,:\,\,\,\sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,{{\left( { - \frac{1}{6}} \right)}^{n - 1}}} = \sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{6^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {\frac{{6{{\left( { - 1} \right)}^{n - 1}}}}{n}} \)\(\displaystyle x = \frac{7}{{24}}\,\,\,:\,\,\,\sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,{{\left( {\frac{1}{6}} \right)}^{n - 1}}} = \sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,\frac{1}{{{6^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {\frac{6}{n}} \)
Now, the first series is an alternating harmonic series which we know converges (or you could just do a quick Alternating Series Test to verify this) and the second series diverges by the \(p\)-series test.
Show Step 6The interval of convergence is below and for summary purposes the radius of convergence is also shown.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\rm{Interval : }}\frac{5}{{24}} \le x < \frac{7}{{24}}}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{1}{{24}}}}\]