Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

In problems 1-6, determine the convergence set of the given power series.
$\sum_{n = 1}^{\infty} \frac{4}{n^{2} + 2 n} {(x - 3)}^{n}$

The set is, $x \in [2, 4]$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:To Find the Radius of convergence

Use the ratio test to determine the radius of convergence.

$\begin{matrix} \lim_{n \to \infty} | \frac{a_{n}}{a_{n + 1}} | = \lim_{n \to \infty} |\frac{\frac{4}{n^{2} + 2 n}}{\frac{4}{{(n + 1)}^{2} + 2 (n + 1)}}| \\ = \lim_{n \to \infty} |\frac{n^{2} + 2 n}{n^{2} + 2 n + 1 + 2 n + 2} \times \frac{4}{4}| \\ = \lim_{n \to \infty} |\frac{n^{2} + 2 n}{n^{2} + 4 n + 3}| \\ = \lim_{n \to \infty} |\frac{n^{2} (1 + (2 / n))}{n^{2} (1 + (4 / n) + (3 / n^{2}))}| \\ = \lim_{n \to \infty} |\frac{(1 + (2 / n))}{(1 + (4 / n) + (3 / n^{2}))}| \\ = 1 \end{matrix}$

The radius of convergence is 1, therefore convergent set for the given power series is .

$| x - 3 | < 1$

Step 2: Find the set of convergence

To completely identify the convergence set, we have to check whether the boundary points 2 and 4 are included in the set or not.

Checking at $x = 2$ , by substituting x by 2,

$\begin{matrix} \sum_{n = 0}^{\infty} \frac{4}{n^{2} + 2 n} {(x - 3)}^{n} = \sum_{n = 0}^{\infty} \frac{4}{n^{2} + 2 n} {(2 - 3)}^{n} \\ = \sum_{n = 0}^{\infty} \frac{4}{n^{2} + 2 n} {(- 1)}^{n} \end{matrix}$

The above series is an alternating harmonic series, which is convergent in nature, thus the point 2 is included in the convergent set.

Similarly, checking at x-4, by substituting x by 4 ,

$\begin{matrix} \sum_{n = 0}^{\infty} \frac{4}{n^{2} + 2 n} {(x - 3)}^{n} = \sum_{n = 0}^{\infty} \frac{4}{n^{2} + 2 n} {(4 - 3)}^{r} \\ = \sum_{n = 0}^{\infty} \frac{4}{n^{2} + 2 n} \end{matrix}$

Since, $n^{2} + 2 n > n^{2} \Rightarrow \frac{1}{n^{2} + 2 n} < \frac{1}{n^{2}}$ and $\sum \frac{1}{n^{2}}$ is convergent, then, by the comparison test it follows that $\sum \frac{1}{n^{2} + 2 n}$ is also convergent.

The convergent set for the given power series is. $x \in [2, 4]$

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In problems 1-6, determine the convergence set of the given power series.
$\sum_{n = 1}^{\infty} \frac{4}{n^{2} + 2 n} {(x - 3)}^{n}$

Step by Step Solution

Step 1:To Find the Radius of convergence

Step 2: Find the set of convergence

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

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Step by Step Solution

Step 1:To Find the Radius of convergence

Step 2: Find the set of convergence

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In problems 1-6, determine the convergence set of the given power series.
$\sum_{n = 1}^{\infty} \frac{4}{n^{2} + 2 n} {(x - 3)}^{n}$