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: 19

Answer

By dividing any latter term by its former we can get the common ration (r). $-2 \div 10 = -0.2$ or $\frac{-1}{5}$ Note: |$\frac{-1}{5}|\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 $10$ and using the calculated r and we get $S= \frac{a}{1-r}$ = $\frac{10}{1-\frac{-1}{5}} = \frac{10}{\frac{6}{5}} = \frac{25}{3}$ Result: the series converges and its sum is $\frac{25}{3}$ .

Work Step by Step

We look at the ratio change from term to term we find that (Same as exercise 17 and 18) : Dividing any latter term by its former we can get the common ration ($r$). $-2 \div 10 = -0.2$ or $\frac{-1}{5}$ . Also, $0.4\div -2 =-0.2$ . Note that |$\frac{-1}{5}|\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 $10$ and using the calculated $r$ and we get $S= \frac{a}{1-r}$ = $\frac{10}{1-\frac{-1}{5}} = \frac{10}{\frac{6}{5}} = \frac{25}{3}$ Result: the series converges and its sum is $\frac{25}{3}$ .
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