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 - Page 756: 15


a) $\lim\limits_{n \to \infty} a_{n}$ = $\lim\limits_{n \to \infty} \frac{2n}{3n+1} = \frac{2}{3}$ ; Convergent . b) $\lim\limits_{n \to \infty} a_{n} = \frac{2}{3}$ ; the series $\Sigma a_{n} $ diverges.

Work Step by Step

Part A requires that we refer to definition in page 736 and take the limit of the sequence. We take n as a common factor from both numerator and denominator and they cancel out each other which leaves us with 2/3 after taking n to infinity. Part B requires that we use the result from part A and state the test for divergence of series and conclude that since the limit of the sequence gives wasn't zero, the series corresponding to the sequence is divergent.
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.