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 - 11.1 Exercises - Page 725: 72

Answer

The sequence is not monotonic and not bounded.

Work Step by Step

$a_{n}=(-2)^{n+1}$ {$a_{n}$} is increasing if $a_{n} \lt a_{n+1}$ for all $n \geq 1$ {$a_{n}$} is decreasing if $a_{n} \gt a_{n+1}$ for all $n \geq 1$ {$a_{n}$} is monotonic if it is either increasing or decreasing. The odd terms of the sequence result in $4,16,64,...$ and are increasing. The even terms of the sequence result in $-8, -32,-128,...$ and are decreasing. Therefore {$a_{n}$} is not monotonic. The terms of the sequence are $4,-8,16,-32,64,...$ By definition, {$a_{n}$} is bounded above if there is a number $M$ such that $a_{n} \leq M$ for all $n \geq 1$ and is bounded below if there is a number $m$ such that $m \leq a_{n}$ for all $n\geq 1$ The odd terms $4,16,64...$ are bounded below by 4 and the even terms $-8,-32,-128,...$ are bounded above by -8. Therefore, the sequence $a_{n}=(-2)^{n+1}$ is not bounded.
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