Answer
$=\dfrac {e^{2}}{e^{2}-1}$
Work Step by Step
$\sum ^{\infty }_{k=1}e^{-2n}=\sum ^{\infty }_{k=1}\left( e^{-2}\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=e^{-2};r=e^{-2}\Rightarrow S_{\infty }=\dfrac {e^{-2}}{1-e^{-2}}=\dfrac {e^{2}}{e^{2}-1}$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
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 - Page 623: 25
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
