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

Answer

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.