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 489: 129

Answer

No. Counterexample: $1,5,1,5,1,5,...$

Work Step by Step

We have seen examples of sequences where terms alternate between two constant values, and we found that such sequences do not converge. An example: $1,5,1,5,1,5,...$ This sequence is bounded from above by 5, but is not convergent. If we set $\epsilon=1/2$, and L is the limit of the sequence, because of the alternating 1's, it must be that $L\in(0.5,1.5),$ and because of the alternating $5$'s, $L\in(4.5,5.5).$ L can not exist because it can't be in both intervals.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.