University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 536: 13


$\dfrac{1}{1+x}=\Sigma_{n=0}^\infty (-1)^n x^n$

Work Step by Step

Since, we know that the Maclaurin Series for $\frac{1}{1-x}$ is defined as: $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n$ We need to replace in the above series $x$ with $-x$. Now, $\dfrac{1}{1-(-x)}=\Sigma_{n=0}^\infty (-x)^n$ Thus, $\dfrac{1}{1+x}=\Sigma_{n=0}^\infty (-1)^n x^n$
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.