Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 9 - Infinite Series - 9.8 Maclaurin And Taylor Series; Power Series - Exercises Set 9.8 - Page 667: 31

Answer

The interval of convergence for the series is: $(-\infty, \infty)$ and the radius of convergence is $R=\infty$.

Work Step by Step

Ratio Test: Let us consider a series $\Sigma a_k$ whose limit $l$ can be obtained as: $ l=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|$ 1. For $l \lt 1$, the series is absolutely convergent. 2. For $l \gt 1$, the series is divergent. 3. For $l = 1$, the series is inconclusive. Therefore, $ L=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|\\=\lim\limits_{k \to \infty} \dfrac{x^{k+1}}{(k+1)!} \times \dfrac{k!}{x^k} \\=\lim\limits_{k \to \infty} \dfrac{|x|}{k+1}\\=0$ We can see that $ l \lt 1$ for all $x$, so the given series converges absolutely for all $x$. Therefore, the interval of convergence for the series is: $(-\infty, \infty)$ and the radius of convergence is $R=\infty$.
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