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 717: 67

Answer

$a_{1} = 0$ and for $n \gt 1$, $a_{n} = \frac{2}{n^{2}+n}$ $\Sigma^{\infty}_{n=1} a_{n}=1$

Work Step by Step

$s_{n} = \frac{n-1}{n+1}$ If $n=1, a_{1}=s_{1} = \frac{1-1}{1+1} = 0$ For $n \gt 1$ (or $n \geq 2$), $a_{n} = s_{n} - s_{n-1}$ $=\frac{n-1}{n+1}-\frac{(n-1)-1}{(n-1)+1}$ $=\frac{n-1}{n+1}-\frac{n-2}{n}$ $=\frac{n(n-1)-(n+1)(n-2)}{n(n+1)}$ $=\frac{n(n-1)-(n+1)(n-2)}{n^{2} + n}$ $=\frac{2}{n^{2}+n}$ $\Sigma^{\infty}_{n=1}a_{n} = \lim\limits_{n \to \infty}s_{n}$ $= \lim\limits_{n \to \infty} \frac{n-1}{n+1}$ $= \lim\limits_{n \to \infty} \frac{n}{n}$ $=1$
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