Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises: 20

Answer

By dividing any latter term by its former we can get the common ration ($r$). $ 0.5\div 2 = 0.25$ or $\frac{1}{4}$ $r$ = |$\frac{1}{4}|\lt1 $ which means that this geometric series will converge. Now, use the formula from definition 4 in page 750 , knowing that $a$ refers to the first term which is $2$ and using the calculated $r$ and we get $S= \frac{a}{1-r}$ = $\frac{2}{1-\frac{1}{4}} = \frac{2}{\frac{3}{4}} = \frac{8}{3}$ Result: the series 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. Now, use the formula from definition 4 in page 750 , knowing that $a$ refers to the first term which is $2$ and using the calculated $r$ and we get $S= \frac{a}{1-r}$ = $\frac{2}{1-\frac{1}{4}} = \frac{2}{\frac{3}{4}} = \frac{8}{3}$ Result: the series converges and its sum is $\frac{8}{3}$ .
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