Multivariable Calculus, 7th Edition

Published by Brooks Cole

Chapter 11 - Infinite Sequences and Series - 11.2 Exercises - Page 735: 5

Answer

The given series is convergent

Work Step by Step

$\Sigma^{\infty}_{n=1} \frac{1}{n^{3}}$ $a_{n} = \frac{1}{n^{3}}$ partial sum $s_{n} = a_{1} + a_{2} +...+a_{n}$ $n=1$ $a_{1}= 1.00$ $s_{1}=1.00$ $n=2$ $a_{2}=0.125$ $s_{2}=1.125$ $n=3$ $a_{3}=0.03704$ $s_{3}=1.1620$ $n=4$ $a_{4}=0.01563$ $s_{4}=1.1777$ $n=5$ $a_{5}=0.008$ $s_{5}=1.1857$ $n=6$ $a_{6}=0.00463$ $s_{6}=1.1903$ $n=7$ $a_{7}=0.00292$ $s_{7}=1.1932$ $n=8$ $a_{8}=0.00195$ $s_{8}=1.1952$ We observe that all the terms of the series are approaching to approximately 1.2 Therefore the given series is convergent.

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