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.
$\sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k}$

The interval of convergence for power series is $(- 4, 0)$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

The given power series is:

$\sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k}$

Step 2. Find the interval of convergence.

Let us assume $b_{k} = \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k}$ therefore

$b_{k + 1} = \frac{{[(k + 1)!]}^{2}}{[2 (k + 1)]!} {(x + 2)}^{k + 1}$

The ratio for the absolute convergence is

$\lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k \to \infty} |\frac{\frac{{[(k + 1)!]}^{2}}{[2 (k + 1)]!} {(x + 2)}^{k + 1}}{\frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k}}| = \lim_{k \to \infty} |\frac{{(k + 1)}^{2}}{(2 k + 2) (2 k + 1)} (x + 2)| = \lim_{k \to \infty} |(x + 2)| \frac{(k + 1)}{(2 k + 1)} \frac{1}{2}$

Here the limit role="math" $\frac{1}{2} |x + 2|$ is So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\frac{1}{2} |x + 2| < 1$ that is $- 4 < x < 0$ .

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 $x = - 4$

$\sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k} = \sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(- 4 + 2)}^{k} = \sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(- 2)}^{k}$

The result is just a constant multiple of the arithmetic series, which diverges conditionally.

So, when $x = 0$

$\sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k} = \sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(0 + 2)}^{k} = \sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(2)}^{k}$

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

Therefore, the interval of convergence of the power series $\sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k}$ is $(- 4, 0)$ .

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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.
$\sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k}$

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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Step by Step Solution

Step 1. Given information.

Step 2. 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.
$\sum_{k = 0}^{\infty} \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k}$