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: 40

Answer

The interval of convergence for the series is: $[-1,1]$ and the radius of convergence is $R=1$.

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)[\ln (k+1)]^2} \times \dfrac{x^k}{k(\ln k)^2} \\=\lim\limits_{k \to \infty} \dfrac{|x|k(\ln k)^2}{(k+1)(\ln (k+1)^2}\\=|x|$ So, we can conclude that the given series converges absolutely if $l=|x| \lt 1$ or $-1 \lt x \lt 1$. The test is inconclusive if $|x| =1$ for $x=1$ or $x=-1$. It diverges if $l=|x| \gt 1$ . Therefore, the interval of convergence for the series is: $[-1,1]$ and the radius of convergence is $R=1$.
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