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: 57

Answer

Series converges and $S_{\infty}= \dfrac{8}{5}$

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 and $a_1$= the first term Now, $r=\dfrac{a_2}{a_1} = \dfrac{-1/2}{2}=\dfrac{-1}{4}$ Since $|\dfrac{-1}{4}|=\dfrac{1}{4}\lt 1$, so the infinite geometric series converges. Next, we will find the sum of the infinite geometric series when $a_1 =2$ and $r=\dfrac{-1}{4}$, $S_{\infty} = \dfrac{a_1}{1-r} = \dfrac{2}{1-\dfrac{-1}{4}}=\dfrac{2}{5/4}$ Therefore, the sum of the convergent infinite geometric series is: $S_{\infty}= \dfrac{8}{5}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.