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 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 618: 13


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

Work Step by Step

Consider the Maclaurin Series for $\dfrac{1}{1-x}$ as follows: $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n$ Plug $-x$ instead of $x$. we get $\dfrac{1}{1-(-x)}=\Sigma_{n=0}^\infty (-x)^n$ This implies that $\dfrac{1}{1+x}=\Sigma_{n=0}^\infty (-1)^n x^n$
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.