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: 11


The series is divergent.

Work Step by Step

$a_{n}=\frac{n}{\sqrt (n^{2}+4)}$ partial sum: $s_{n} = a_{1} + a_{2}+...+a_{n}$ $a_{n}$ (blue dots) $s_{n}$ (red dots} $n=1$ $s_{1}=0.4472$ $a_{1}=0.4472$ $n=2$ $s_{2}=1.1543$ $a_{2}=0.7071$ $n=3$ $s_{3}=1.9864$ $a_{3}=0.8321$ $n=4$ $s_{4}=2.8808$ $a_{4}=0.8944$ $n=5$ $s_{5}=3.0893$ $a_{5}=0.9285$ $n=6$ $s_{6}=4.7580$ $a_{6}=0.9487$ $n=7$ $s_{7}=5.7195$ $a_{7}=0.9615$ $n=8$ $s_{8}=6.6896$ $a_{8}=0.9701$ $n=9$ $s_{9}=7.6658$ $a_{9}=0.9762$ $n=10$ $s_{10}=8.6464$ $a_{10}=0.9806$ The sequence terms $a_{n}$ appear to approach 1 Check limit $L$ $L = \lim\limits_{n \to \infty} \frac{n}{\sqrt (n^{2}+4)}$ $L^{2} = \lim\limits_{n \to \infty} \frac{n^{2}}{n^{2}+4} \times \frac{\frac{1}{n^{2}}}{\frac{1}{n^{2}}}$ $L^{2}=\lim\limits_{n \to \infty} \frac{1}{1+\frac{4}{n^{2}}} = \frac{1}{1+0} = 1$ $L=1 \ne 0$ The limit is not zero, so the series is divergent
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