College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.3 - Geometric Sequences - 8.3 Exercises - Page 615: 70


convergent; $\S_{\infty}=\dfrac{1}{648}$

Work Step by Step

RECALL: (1) The sum of an infinite geometric series is convergent if $|r| < 1$. The sum is given by the formula: $S_{\infty}=\dfrac{a}{1-r}$ (2) An infinite geometric series is divergent if $|r|>1$. (3) The common ratio $r$ of a geometric series can be found by dividing any term by the term before it: $r = \dfrac{a_n}{a_{n-1}}$ Solve for $r$ to obtain: $\require{cancel} r = \dfrac{\frac{1}{3^8}}{\frac{1}{3^6}} \\r=\dfrac{1}{3^8} \cdot \dfrac{3^6}{1} \\r=\dfrac{3^6}{3^8} \\r=\dfrac{\cancel{3^6}}{\cancel{3^8}3^2} \\r=\dfrac{1}{3^2}=\dfrac{1}{9}$ Since $|\frac{1}{9}|<1$, then the series converges. Solve for the sum using the formula above, with $a=\dfrac{1}{3^6}$ and $r=\frac{1}{9}$, to obtain: $\require{cancel} S_{\infty}=\dfrac{\frac{1}{3^6}}{1-\frac{1}{9}} \\S_{\infty}=\dfrac{\frac{1}{3^6}}{\frac{9}{9}-\frac{1}{9}} \\S_{\infty}=\dfrac{\frac{1}{3^6}}{\frac{8}{9}} \\S_{\infty}=\dfrac{\frac{1}{3^6}}{\frac{8}{3^2}} \\S_{\infty} = \dfrac{1}{3^6} \cdot \frac{3^2}{8} \\S_{\infty} = \dfrac{1}{\cancel{3^6}3^4} \cdot \dfrac{\cancel{3^2}}{8} \\S_{\infty}=\dfrac{1}{648}$
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