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: 22

Answer

The geometric series is divergent

Work Step by Step

$\Sigma^{\infty}_{n=1}ar^{n-1} = a+ar+ar^{2}$ If $|r| \lt 1$ the geometric series is convergent $\Sigma^{\infty}_{n=1}ar^{n-1} = \frac{a}{1-r}$ $|r| \lt 1$ If $|r| \geq 1$ the geometric series is divergent $\Sigma^{\infty}_{n=1} \frac{10^{n}}{(-9)^{n-1}}$ rewrite $n$-th term of the geometric series in the form $ar^{n-1}$ $=\Sigma^{\infty}_{n=1}10\frac{10^{n-1}}{(-9)^{n-1}}$ $=\Sigma^{\infty}_{n=1}10(\frac{10}{-9})^{n-1}$ $=\Sigma^{\infty}_{n=1}10(-\frac{10}{9})^{n-1}$ First term, $a=10$ and the common ratio is $r= -\frac{10}{9}$ $|r|=|-\frac{10}{9}|$ $=\frac{10}{9}$ $=1.1111$ $\gt 1$ Since $|r| \gt 1$, the geometric series is divergent.
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