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: 53


$S_{\infty}= \dfrac{3}{2}$

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 and $a_1$= the first term Now, $r=\dfrac{a_2}{a_1} = \dfrac{(1/3)}{1}=\dfrac{1}{3}$ Since $\dfrac{1}{3}|\lt 1$, so the infinite geometric series converges. Next, we will find the sum of the infinite geometric series when $a_1 = 1$ and $r=\dfrac{1}{3}$, $S_{\infty} = \dfrac{a_1}{1-r} = \dfrac{1}{1-\dfrac{1}{3}} \\ =\dfrac{1}{\dfrac{2}{3}} \\ = \dfrac{3}{2} $ Therefore, the sum of the convergent infinite geometric series is: $S_{\infty}= \dfrac{3}{2}$
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