University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 46



Work Step by Step

Here, we have: $ 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}}}$ Thus, we have the sum: $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [\dfrac{1}{2}-\dfrac{1}{2^{\frac{1}{n+1}}}]=\lim\limits_{n \to \infty} \dfrac{1}{2}- \lim\limits_{n \to \infty}\dfrac{1}{2^{\frac{1}{n+1}}}=\dfrac{1}{2}-1=-\dfrac{1}{2}$
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