Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 61


The series converges if $x ϵ (-\infty, -2)$ U $(2,\infty)$ and the sum is $\frac{x}{x-2}$

Work Step by Step

Given:$\Sigma^{\infty}_{n=0} \frac{2^{n}}{x^{n}} $ $\Sigma^{\infty}_{n=0} \frac{2^{n}}{x^{n}} = \Sigma^{\infty}_{n=0}(\frac{2}{n})^{n}$ Here, $a=1$ and $r=\frac{2}{x}$ The series converges when $|r| \lt 1$ $|\frac{2}{x}| \lt 1$ $-1 \lt \frac{2}{x} \lt 1$ $-1 \lt \frac{2}{x}$ or $\frac{2}{x} \lt 1$ $x \lt -2$ or $x \gt 2$ The series converges if $x ϵ (-\infty, -2)$ U $(2,\infty)$ Sum can be calculated as follows: $\Sigma^{\infty}_{n=0} \frac{2^{n}}{x^{n}} = \frac{a}{1-r}$ $=\frac{1}{1-\frac{2}{x}}$ $=\frac{x}{x-2}$
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