Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 4

Answer

The Taylor series is $\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{3^{n}(n+1)}(x-4)^{n}$ and $R=3$

Work Step by Step

Use the given value of $f^{n}(4)$ in the formula for the Taylor series at $x=4$, then simplify $\Sigma_{n=0}^{\infty}\frac{(-1)^{n}n!}{n!3^{n}(n+1)}(x-4)^{n}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{3^{n}(n+1)}(x-4)^{n}$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=|\frac{x-4}{3}|\lt 1$ Hence, the Taylor series is $\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{3^{n}(n+1)}(x-4)^{n}$ and $R=3$
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.