Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.3 Infinite Series - 8.3 Exercises: 38

Answer

$-\dfrac {1}{e+1}$

Work Step by Step

$\sum ^{\infty }_{n=1}\left( -e\right) ^{-k}=\sum ^{\infty }_{k=1}\left( -\dfrac {1}{e}\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=\left( -\dfrac {1}{e}\right) ^{1}=-\dfrac {1}{e};r=-\dfrac {1}{e}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {-\dfrac {1}{e}}{1-\left( -\dfrac {1}{e}\right) }=-\dfrac {1}{e+1}$
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