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: 25

Answer

$=\dfrac {e^{2}}{e^{2}-1}$

Work Step by Step

$\sum ^{\infty }_{k=1}e^{-2n}=\sum ^{\infty }_{k=1}\left( e^{-2}\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=e^{-2};r=e^{-2}\Rightarrow S_{\infty }=\dfrac {e^{-2}}{1-e^{-2}}=\dfrac {e^{2}}{e^{2}-1}$
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