Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.4 - Comparison Tests - Exercises 10.4 - Page 591: 6



Work Step by Step

Let $u_n=\dfrac{1}{n3^{n}}$ When the numerator increases, the value of fraction is always increasing and also, when the denominator decreases, the value of fraction is always increasing. $\implies a_n \leq \dfrac{1}{3^{n}}$ we can see that the series $\Sigma_{n=1}^\infty \dfrac{1}{3^n}$ shows a convergent geometric series and has common ratio $r=\dfrac{1}{3} \gt 1$ . Hence, the series converges by the comparison test.
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.