University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 70


converges to $\dfrac{1}{1+x^2}$

Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ The given series $\sum_{n=0}^\infty (-1)^n x^{2n}=\sum_{n=0}^\infty (-x^2)^n$ shows a convergent geometric series with first term, $a=1$ and common ratio $r =-x^2$ $S=\dfrac{1}{1-(-x^2)}=\dfrac{1}{1+x^2}$ Hence, the given series converges to $\dfrac{1}{1+x^2}$ for $|x| \lt 1$
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