Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.8 Power Series - 11.8 Exercises - Page 791: 29


(a) Yes, it is convergent (b) No, it does not follow that $\Sigma_{n=0}^{\infty}c_{n}(-4)^{n}$ converges.

Work Step by Step

(a) For a power series there is a positive number $R$ such that the series converges when $|x-a|\lt R$ In the given problem, $|x-a|=4$ Thus, $R\gt 4$ and the minimum interval of convergence would be $(a-4,a+4)$ Since, $|-2|=2\lt 4$ , that is within the interval of convergence for the minimum $R=4$ and it follows that $\Sigma_{n=0}^{\infty}c_{n}(-2)^{n}$ converges also. (b) The given function could be either converge or diverge Thus, It does not follow that $\Sigma_{n=0}^{\infty}c_{n}(-4)^{n}$ 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.