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 - 14.3 Geometric Sequences and Series - 14.3 Exercise Set - Page 912: 52

Answer

Yes, the given infinite geometric series has a limit, and the value of the limit is ${{S}_{\infty }}=\frac{37}{99}$.

Work Step by Step

The infinite geometry series is: $0.37+0.0037+0.000037+\cdots $ Here ${{a}_{1}}=0.37,{{a}_{2}}=0.0037$. The value of $\left| r \right|$ is: $\begin{align} & \left| r \right|=\left| \frac{{{a}_{2}}}{{{a}_{1}}} \right| \\ & =\left| \frac{0.0037}{0.37} \right| \\ & =\left| 0.01 \right| \\ & =0.01 \end{align}$ Find the limit of the infinite geometric series for the formula ${{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$. $\begin{align} & {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r} \\ & =\frac{0.37}{1-0.01} \\ & =\frac{0.37}{0.99} \\ & =\frac{37}{99} \end{align}$ Thus, the limit of the infinite geometric series is ${{S}_{\infty }}=\frac{37}{99}$.
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