Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 716: 20

Answer

Converges and its sum is $\frac{8}{3}$

Work Step by Step

We look at the ratio change from term to term and we find that: By dividing any latter term by its former we can get the common ration ($r$). $ 0.5\div 2 = 0.25$ or $\frac{1}{4}$ . Also, $0.125\div 0.5 = 0.25$ . Note that |$\frac{1}{4}|\lt1 $ which means that this geometric series will converge. $a$ refers to the first term which is $2$ and using the calculated $r$ we get $S= \frac{a}{1-r}$ = $\frac{2}{1-\frac{1}{4}} = \frac{2}{\frac{3}{4}} = \frac{8}{3}$ Hence, the series converges and its sum is $\frac{8}{3}$
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