Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 715: 4

Answer

$\Sigma^{\infty}_{n=1}a_{n}=\frac{1}{4}$

Work Step by Step

$s_{n} = \frac{n^{2}-1}{4n^{2}+1}$ $\lim\limits_{n \to \infty}s_{n} = \lim\limits_{n \to \infty} (\frac{n^{2}-1}{4n^{2}+1})$ $\lim\limits_{n \to \infty}(\frac{1-\frac{1}{n^{2}}}{4+\frac{1}{n^{2}}})$ $=\frac{1}{4}$ $\Sigma^{\infty}_{n=1}a_{n}=\frac{1}{4}$
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