University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 25


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

Work Step by Step

We know that the Maclaurin Series for $e^x$ and $\dfrac{1}{1+x}$ is defined as: $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$ and $\dfrac{1}{1+x}=1-x+x^2-x^3+x^4+...$ Then $e^x+\dfrac{1}{1+x}=[1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...]+[1-x+x^2-x^3+x^4+...]$ or, $=(1+1)x^0+(1-1)x^1+(\dfrac{1}{2!}+1)x^2+...$ or, $=\Sigma_{n=0}^\infty (\dfrac{1}{n!}+(-1)^n)x^n$
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