Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.2 Exercises - Page 735: 45


The series is convergent and the sum is $\frac{11}{6}$

Work Step by Step

$\Sigma^{\infty}_{n=1} \frac{3}{n(n+3)}$ $S_{n} =\Sigma^{n}_{k=1} \frac{3}{k(k+3)}$ $=\frac{3}{1 \times 4}+ \frac{3}{2 \times 5} + ... +\frac{3}{n(n+3)}$ $S_{n} = \Sigma^{n}_{k=1} \frac{3}{k(k+3)}$ $=\Sigma^{n}_{k=1} (\frac{1}{k} - \frac{1}{k+3})$ $=(1-\frac{1}{4}) + (\frac{1}{2}-\frac{1}{5})+(\frac{1}{3}-\frac{1}{6})+(\frac{1}{4}-\frac{1}{7})+...+(\frac{1}{n} - \frac{1}{n+3})$ $=1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+3}$ $\lim\limits_{n \to \infty}S_{n} = \lim\limits_{n \to \infty}(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+3})$ $=1+\frac{1}{2}+\frac{1}{3}-\frac{1}{\infty+3}$ $=1+\frac{1}{2}+\frac{1}{3}-\frac{1}{\infty}$ $=1+\frac{1}{2}+\frac{1}{3}-0$ $=\frac{6+3+2}{6}$ $=\frac{11}{6}$
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