Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 33


$coshx\geq 1+\frac{x^{2}}{2}$

Work Step by Step

As we know, $$coshx=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}$$ This series converges absolutely for $x$ , Since, $\lim\limits_{n \to \infty}|\frac{\frac{x^{2n+2}}{(2n+2)!}}{\frac{x^{2n}}{(2n)!}}|=\lim\limits_{n \to \infty}|\frac{x^{2}}{(2n+1)(2n+2)}|=0\lt1 $ and thus, by ratio test the series absolutely converges for $x$ Hence, $coshx=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}=1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+...\geq 1+\frac{x^{2}}{2!}$ Thus, $coshx\geq 1+\frac{x^{2}}{2}$
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