Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 716: 47

Answer

The series is convergent and the sum is $e-1$.

Work Step by Step

$\Sigma^{\infty}_{n=1} (e^{\frac{1}{n}}-e^{\frac{1}{n+1}})$ $S_{n}=\Sigma^{n}_{k=1} (e^{\frac{1}{k}}-e^{\frac{1}{k+1}})$ $=e^{1} - e^{\frac{1}{2}} + e^{\frac{1}{2}} - e^{\frac{1}{3}} + e^{\frac{1}{3}} - e^{\frac{1}{4}}+...+e^{\frac{1}{n}}-e^{\frac{1}{n+1}}$ $=e^{1}-e^{\frac{1}{n+1}}$ $\lim\limits_{n \to \infty}S_{n} = \lim\limits_{n \to \infty}(e^{1}-e^{\frac{1}{n+1}})$ $=e-e^{\frac{1}{\infty}}$ $=e-e^{0}$ $=e-1$ The series is convergent and the sum is $e-1$.
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