Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.5 The Ratio and Root Tests and Strategies for Choosing Tests - Exercises - Page 568: 30


The series $\Sigma_{n=1}^{\infty}2^na_n$ converges.

Work Step by Step

Let $b_n=2^na_n$, then applying the ratio test, we have $$ \rho=\lim _{n \rightarrow \infty}\left|\frac{b_{n+1}}{b_{n}}\right|=\lim _{n \rightarrow \infty} \frac{2^{n+1}}{2^n}\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{2}{3}\lt 1 $$ Then the series $\Sigma_{n=1}^{\infty}2^na_n$ converges.
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.