Calculus II - Power Series

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Section 10.14 : Power Series

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4. For the following power series determine the interval and radius of convergence.

$\sum\limits_{n = 0}^\infty {\frac{{{4^{1 + 2n}}}}{{{5^{n + 1}}}}{{\left( {x + 3} \right)}^n}}$

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Start Solution

Okay, 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 2

So, 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 3

So, 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 4

Now, 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 5

To 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 6

The interval of convergence is below and for summary purposes the radius of convergence is also shown.

$\require{bbox} \bbox[2pt,border:1px solid black]{{{\rm{Interval : }} - \frac{{53}}{{16}} < x < - \frac{{43}}{{16}}}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{5}{{16}}}}$