Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 9 - Power Series - 9.2 Properties of Power Series - 9.2 Exercises - Page 683: 19

Answer

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

Work Step by Step

Root Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty} \sqrt[k] a_k$ 1) When $0 \leq l \lt 1$. then the series $\Sigma a_k$ converges 2) When $l \gt 0$. then series $\Sigma a_k$ diverges. 3) When $l = 1$. then series $\Sigma a_k$ may converge or diverge, that is, the test is inconclusive. Here, we have $a_k=(\dfrac{x^k}{k^k})$ Now, $l=\lim\limits_{k \to \infty} \sqrt[k] {(\dfrac{x^k}{k^k})}=|x| \lim\limits_{k \to \infty} (\dfrac{1}{k})^k=0$. So, the radius of convergence for the power series is $R=\dfrac{1}{l}=\dfrac{1}{0}=\infty$ and the interval of convergence is $(-\infty, \infty)$.
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