Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 14 - Sequences, Series, and the Binomial Theorem - Review Exercises: Chapter 14 - Page 927: 6


True statement

Work Step by Step

$10-5+\frac{5}{2}-\ldots $ The limit of the infinite geometric series is given by ${{s}_{\infty }}=\frac{{{a}_{1}}}{1-r}$ When $\left| r \right|<1$ then the limit exists and if $\left| r \right|\ge 1$, no limit exists. To check the limit find the common ratio $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Put $n=1$ in $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$, $\begin{align} & r=\frac{{{a}_{1+1}}}{{{a}_{1}}} \\ & =\frac{{{a}_{2}}}{{{a}_{1}}} \end{align}$ Put ${{a}_{1}}=10$ and ${{a}_{2}}=-5$ in $r=\frac{{{a}_{2}}}{{{a}_{1}}}$ $\begin{align} & r=\frac{{{a}_{2}}}{{{a}_{1}}} \\ & =\frac{-5}{10} \\ & =-\frac{1}{2} \end{align}$ And, $\begin{align} & \left| r \right|=\left| -\frac{1}{2} \right| \\ & =\frac{1}{2} \end{align}$ Thus, $\left| r \right|=\frac{1}{2}<1$, then the limit exists. Therefore, the given statement is true.
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