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: 41


Interval of Convergence: $-\dfrac{1}{3} \lt x \lt \dfrac{1}{3}$ and $\Sigma_{n=1}^{\infty} (3x)^n=\dfrac{1}{1-3x}$

Work Step by Step

We have $\Sigma_{n=1}^{\infty} 3^nx^n= \Sigma_{n=0}^{\infty} (f(x))^n$ The series will converge when $|f(x)|=|3x| \lt 1$ So, $|x| \lt \dfrac{1}{3} \implies -\dfrac{1}{3} \lt x \lt \dfrac{1}{3}$ Now, the sum of a given geometric series is: $S=\dfrac{a}{1-r}=\dfrac{1}{1-3x}$ Thus, the Interval of Convergence is: $-\dfrac{1}{3} \lt x \lt \dfrac{1}{3}$ and $\Sigma_{n=1}^{\infty} (3x)^n=\dfrac{1}{1-3x}$
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