Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.2 Exercises - Page 735: 24


The series is convergent and the sum of the series is $\sqrt 2 +2$.

Work Step by Step

$\Sigma^{\infty}_{n=0} \frac{1}{(\sqrt 2)^{n}} =\Sigma^{\infty}_{n=0} (\frac{1}{\sqrt 2})^{n}$ $a=1$ and $r=\frac{1}{\sqrt 2} \lt 1$ Since $|r| \lt 1$ The series is convergent $=\frac{a}{1-r}$ $=\frac{\sqrt 2}{\sqrt 2 -1}$ $=\sqrt 2 +2$
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