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: 18

Answer

We find common ratio $r$ by dividing any latter term by its former we can get the common ration (r). $\frac{9}{4} \div 3 = \frac{3}{4}$ . So , $r = \frac{3}{4}$ |$\frac{3}{4}|\lt1 $ , meaning that this geometric series will converge. By the formula from definition 4 in page 750 , with $a$ being $4$ and using the calculated r and we get $S= \frac{a}{1-r}$ = $\frac{4}{1-\frac{3}{4}} = \frac{4}{\frac{1}{4}} = 16$ Result: the series converges and its sum is 16 .

Work Step by Step

We look at the ratio change from term to term we find that (Same as exercise 17) : Dividing any latter term by its former we can get the common ration (r). $\frac{9}{4} \div 3 = \frac{3}{4}$ . Also, $3\div 4 = \frac{3}{4}$ . Note that |$\frac{3}{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 $4$ and using the calculated r and we get $S= \frac{a}{1-r}$ = $\frac{4}{1-\frac{3}{4}} = \frac{4}{\frac{1}{4}} = 16$ Result: the series converges and its sum is 16 .
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