Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.7 - Power Series - Exercises 10.7 - Page 613: 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

Remember that the series will converge when $|f(x)|=|\dfrac{\sqrt{x}}{2}-1| \lt 1$ This implies that $-1 \lt \dfrac{\sqrt x}{2}-1 \lt 1$ and $0 \lt \dfrac{\sqrt x}{2} \lt 2$ or, $0 \lt x \lt 16$ This means that the interval of convergence is : $0 \lt x \lt 16$ Next, we will find the sum of the given geometric series $S=\dfrac{a}{1-r}=\dfrac{1}{1-(\dfrac{\sqrt{x}}{2}-1)}=\dfrac{2}{4-\sqrt x}$ and $\Sigma_{n=0}^{\infty} (\dfrac{\sqrt{x}}{2}-1)^n=\dfrac{2}{4-\sqrt x}$
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