Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Find the interval of convergence for each power series in Exercises 21–48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.
$\sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k}^{}$

The interval of convergence for power series is $(\frac{1}{2}, \frac{5}{2})$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

The given power series is:

$\sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k}$

Step 2. Find the interval of convergence.

Let us assume $b_{k} = {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k}$ therefore

$b_{k + 1} = {(- 1)}^{k + 1} \frac{\sqrt{k + 1}}{{(k + 1)}^{3}} {(x - \frac{3}{2})}^{k + 1}$

The ratio for the absolute convergence is

$\lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k \to \infty} |\frac{{(- 1)}^{k + 1} \frac{\sqrt{k + 1}}{{(k + 1)}^{3}} {(x - \frac{3}{2})}^{k + 1}}{{(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k}}| = \lim_{k \to \infty} |(- 1) \frac{k^{3}}{{(k + 1)}^{3}} \frac{\sqrt{k + 1}}{\sqrt{k}} (x - \frac{3}{2})| = \lim_{k \to \infty} |x - \frac{3}{2}| {(\frac{k}{k + 1})}^{3} \sqrt{\frac{k + 1}{k}}$

Here the limit is $|x - \frac{3}{2}|$ So, by the ratio test of absolute convergence, we know that series will converge absolutely when $|x - \frac{3}{2}| < 1$ that is $\frac{1}{2} < x < \frac{5}{2}$ .

Step 3. Find the interval of convergence.

Now, since the intervals are finite so we analyze the behavior of the series at the endpoints

So, when $x = \frac{1}{2}$

$\sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k} = \sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(\frac{1}{2} - \frac{3}{2})}^{k} = \sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(- 1)}^{k} = \sum_{k = 1}^{\infty} \frac{\sqrt{k}}{k^{3}} {(- 1)}^{2 k}$

The result is the alternating multiple of the harmonic series, which diverges.

So, when $x = \frac{5}{2}$

$\sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k} = \sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(\frac{5}{2} - \frac{3}{2})}^{k} = \sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(1)}^{k} = \sum_{k = 1}^{\infty} \frac{\sqrt{k}}{k^{3}} {(- 1)}^{k}$ The result is just a constant multiple of the harmonic series which converges conditionally.

Therefore, the interval of convergence of the power series is $\sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k} is (\frac{1}{2}, \frac{5}{2})$ .

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Calculus

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Short Answer

Find the interval of convergence for each power series in Exercises 21–48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.
$\sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k}^{}$

Step by Step Solution

Step 1. Given information.

Step 2. Find the interval of convergence.

Step 3. Find the interval of convergence.

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Calculus

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Short Answer

Find the interval of convergence for each power series in Exercises 21–48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.∑k=1∞-1kkk3x-32k

Step by Step Solution

Step 1. Given information.

Step 2. Find the interval of convergence.

Step 3. Find the interval of convergence.

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Find the interval of convergence for each power series in Exercises 21–48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.
$\sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k}^{}$