Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 744: 11


$2, \frac{2}{3},\frac{2}{5},\frac{2}{7},\frac{2}{9}$

Work Step by Step

Given: $a_{1}=2,a_{(n+1)}=\frac{a_{n}}{1+a_n}$ $a_{1}=6$ $a_{1+1}=a_{(2)}=\frac{a_{1}}{1+a_1}=\frac{2}{1+2}=\frac{2}{3}$ $a_{2+1}=a_{(3)}=\frac{a_{2}}{1+a_2}=\frac{2/3}{1+\frac{2}{3}}=\frac{2}{5}$ $a_{3+1}=a_{(4)}=\frac{a_{3}}{1+a_3}=\frac{2/5}{1+\frac{2}{5}}=\frac{2}{7}$ $a_{4+1}=a_{(5)}=\frac{a_{4}}{1+a_4}=\frac{2/7}{1+\frac{2}{7}}=\frac{2}{9}$ Hence we see that the first five terms are $2, \frac{2}{3},\frac{2}{5},\frac{2}{7},\frac{2}{9}$.
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