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: 45

Answer

Interval of Convergence : $0 \lt x \lt 16$ and $\Sigma_{n=0}^{\infty} (\dfrac{\sqrt{x}}{2}-1)^n=\dfrac{2}{4-\sqrt x}$

Work Step by Step

The series will converge when $|f(x)|=|\dfrac{\sqrt{x}}{2}-1| \lt 1$ So, $-1 \lt \dfrac{\sqrt x}{2}-1 \lt 1$ $ \implies 0 \lt \dfrac{\sqrt x}{2} \lt 2$ $ \implies 0 \lt x \lt 16$ Now, the sum of the given geometric series is: $S=\dfrac{a}{1-r}=\dfrac{1}{1-(\dfrac{\sqrt{x}}{2}-1)}=\dfrac{2}{4-\sqrt x}$ Thus, the Interval of Convergence is: $0 \lt x \lt 16$ and $\Sigma_{n=0}^{\infty} (\dfrac{\sqrt{x}}{2}-1)^n=\dfrac{2}{4-\sqrt x}$
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