Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.2 - Infinite Series - Exercises 10.2 - Page 580: 46

Answer

$-\dfrac{1}{2}$

Work Step by Step

The nth partial sums are: $ s_n=(\dfrac{1}{2^{\frac{1}{1}}}-\dfrac{1}{2^{\frac{1}{2}}})+(\dfrac{1}{2^{\frac{1}{2}}}-\dfrac{1}{2^{\frac{1}{3}}})+.....(\dfrac{1}{2^{\frac{1}{n}}}-\dfrac{1}{2^{\frac{1}{n+1}}})=[\dfrac{1}{2}-\dfrac{1}{2^{\frac{1}{n+1}}}]$ Now, we have $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [\dfrac{1}{2}-\dfrac{1}{2^{\frac{1}{n+1}}}]$ or, $=\lim\limits_{n \to \infty} \dfrac{1}{2}- \lim\limits_{n \to \infty}\dfrac{1}{2^{\frac{1}{n+1}}}$ or, $=\dfrac{1}{2}-1$ or, $=-\dfrac{1}{2}$
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