Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.3 Geometric Sequences; Geometric Series - 11.3 Assess Your Understanding - Page 845: 65

Answer

Series Converges. and $S_{\infty}=\dfrac{18}{5}$

Work Step by Step

The common ratio of a geometric sequence is equal to the quotient of any term and the term before it: $ \ r = \dfrac{a_n}{a_{n-1}}$ or, $r=\dfrac{a_2}{a_1}$ The sum of a convergent infinite geometric series is given by the formula: $S_{\infty}=\dfrac{a_1}{1-r}$ and a geometric series converges if $|r| \lt 1$. where $r \ =common \ ratio \ =\dfrac{-2}{3}$ Since $r=|\dfrac{-2}{3}|=\dfrac{2}{3} \lt 1$, so the infinite geometric series Converges. First term: $a_1=6(\dfrac{-2}{3})^{1-1}=(6)(1)=6$ Next, $S_{\infty}=\dfrac{6}{1-(\dfrac{-2}{3})}=\dfrac{6}{5/3}$ Therefore, the sum of the convergent infinite geometric series is: $S_{\infty}=\dfrac{18}{5}$
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