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: 46


The series is convergent and the sum is $cos(1)-1$.

Work Step by Step

$\Sigma^{\infty}_{n=1} (cos\frac{1}{n^{2}}-cos\frac{1}{(n+1)^{2}})$ $S_{n}=\Sigma^{n}_{k=1} (cos\frac{1}{n^{2}}-cos\frac{1}{(n+1)^{2}})$ $=(cos\frac{1}{1^{2}}-cos\frac{1}{(1+1)^{2}})+(cos\frac{1}{2^{2}}-cos\frac{1}{(2+1)^{2}})+(cos\frac{1}{3^{2}}-cos\frac{1}{(3+1)^{2}})+...+(cos\frac{1}{n^{2}}-cos\frac{1}{(n+1)^{2}})$ $=cos(1)-cos\frac{1}{(n+1)^{2}}$ $\lim\limits_{n \to \infty}S_{n} = \lim\limits_{n \to \infty} (cos(1)-cos\frac{1}{(n+1)^{2}})$ $=cos(1) - cos(\frac{1}{\infty})$ $=cos(1) -cos(0)$ $=cos(1)-1$ Therefore the series is convergent and the sum is $cos(1)-1$
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