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: 77

Answer

{$a_{n}$} is a decreasing sequence and is bounded above.

Work Step by Step

$a_{n} = \frac{n}{n^{2}+1}$ $f(x)=\frac{x}{x^{2}+1}$ $f'(x) = \frac{(x^{2}+1)(1)-x(2x)}{(x^{2}+1)^{2}}$ $=\frac{1-x^{2}}{(x^{2}+1)^{2}} \lt 0$ when $x \gt 1$ Therefore $f$ is decreasing on $(1, \infty)$ and so {$a_{n}$} is a decreasing sequence. {$a_{n}$} is decreasing for $n \geq 1$ $a_{n} \leq a_{1}$ $ \frac{n}{n^{2}+1} \leq \frac{1}{2}$ $a_{n} \leq \frac{1}{2}$ Thus {$a_{n}$} is bounded above.
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