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: 40


The series is divergent.

Work Step by Step

$\Sigma^{\infty}_{n=1}(\frac{3}{5^{n}} + \frac{2}{n})$ $=\Sigma^{\infty}_{n=1} \frac{3}{5^{n}} + \Sigma^{\infty}_{n=1} \frac{2}{n}$ $= 3\Sigma^{\infty}_{n=1} \frac{1}{5^{n}} + 2\Sigma^{\infty}_{n=1} \frac{1}{n}$ First $\Sigma^{\infty}_{n=1} \frac{1}{5^{n}} = \frac{1}{5} + \frac{1}{5^{2}}+\frac{1}{5^{3}}+\frac{1}{5^{4}}+...$ The common ratio $r= \frac{1}{5} \lt 1$ Therefore $\Sigma^{\infty}_{n=1} \frac{1}{5^{n}}$ is convergent. Second $\Sigma^{\infty}_{n=1} \frac{1}{n}$ $s_{1} =1$ $s_{2} =1 + \frac{1}{2}$ $s_{3}= 1 + \frac{1}{2} + (\frac{1}{3}+\frac{1}{4})$ $\gt 1+\frac{1}{2} + (\frac{1}{4}+\frac{1}{4})$ $=1+\frac{2}{2}$ $s_{4}=1 + \frac{1}{2} + (\frac{1}{3}+\frac{1}{4}) + (\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})$ $\gt 1 + \frac{1}{2} + (\frac{1}{4}+\frac{1}{4}) + (\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})$ $=1+\frac{1}{2} +\frac{1}{2} +\frac{1}{2}$ $=1+\frac{3}{2}$ $S_{2n} \gt 1+\frac{n}{2}$ Which shows that $S_{2n}ā†’\infty$ as $nā†’\infty$ Therefore the series $\Sigma^{\infty}_{n=1}\frac{1}{n}$ is divergent. Since the first series is convergent and the second is divergent, the sum of the series is divergent.
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