Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.1 An Overview - 8.1 Exercises - Page 606: 63


a. $\left\{\frac{2}{3},\frac{4}{5},\frac{6}{7},\frac{8}{9}\right\}$ b.$S_n=\frac{2n}{2n+1}$ c. 1

Work Step by Step

a. $\frac{2}{(2(1)-1)(2(1)+1)}=\frac{2}{3}$ $\frac{2}{(2(2)-1)(2(2)+1)}=\frac{2}{15}$ $\frac{2}{15}+\frac{2}{3}=\frac{12}{15}=\frac{4}{5}$ $\frac{2}{(2(3)-1)(2(3)+1)}=\frac{2}{35}$ $\frac{2}{35}+\frac{4}{5}=\frac{30}{35}=\frac{6}{7}$ $\frac{2}{(2(4)-1)(2(4)+1)}=\frac{2}{63}$ $\frac{2}{63}+\frac{6}{7}=\frac{56}{63}=\frac{8}{9}$ b. By looking at the $S_n$ values, one notices each time $n$ increases by one, both the numerator and denominator increase by 2. The numerator is always $2x$ the $n$ value, and the denominator is one greater than the numerator. Therefore, $S_n=\frac{2n}{2n+1}$ 3. The value of the series is $\lim\limits_{n \to \infty}S_n$, which is $\lim\limits_{n \to \infty}\frac{2n}{2n+1}$, which equals 1 because the numerator and denominator have the same order and their leading coefficients are the same.
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