University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 487: 10

Answer

$-2,\ -1,\ -2/3,\ -1/2,\ -2/5,\ -1/3,\ -2/7,\ -1/4,\ -2/9,\ -1/5$

Work Step by Step

$\left[\begin{array}{ll} n & a_{n}\\ \\ 1 & -2\\ \\ 2 & \frac{1}{2}(-2)=-1\\ \\ 3 & \frac{2}{3}(-1)= -\frac{2}{3} \\ \\ 4 & \frac{3}{4}(-\frac{2}{3})= -\frac{1}{2} \\ \\ 5 & \frac{4}{5}(-\frac{1}{2})= -\frac{2}{5} \\ \\ 6 & \frac{5}{6}(-\frac{2}{5})= -\frac{1}{3} \\ \\ 7 & \frac{6}{7}(-\frac{1}{3})= -\frac{2}{7} \\ \\ 8 & \frac{7}{8}(-\frac{2}{7})= -\frac{1}{4} \ \\ 9 & \frac{8}{9}(-\frac{1}{4})= -\frac{2}{9}\\ \\ 10 & \frac{9}{10}(-\frac{2}{9})= -\frac{1}{5} \end{array}\right]$
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