University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.7 - Power Series - Exercises - Page 531: 37


Radius of convergence is $3$

Work Step by Step

We are given that:$a_n=\dfrac{n!}{3 \cdot 6.....3n}x^n$ We need to apply the Ratio Test to the series. $\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty} \dfrac{a_n (x/3)}{a_n}$ or, $=\lim\limits_{n \to \infty} |\dfrac{x}{3}|$ Now, $|\dfrac{x}{3}| \lt 1 \implies |x| \lt 3$ Thus, the radius of convergence is $3$.
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