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: 31

Answer

$=\dfrac {\pi }{\pi -e}$

Work Step by Step

$1+\dfrac {e}{\pi }+\dfrac {e^{2}}{\pi ^{2}}+\dfrac {e^{3}}{\pi ^{3}}-..=\dfrac {a_{1}}{1-r};a_{1}=1;r=\dfrac {e}{\pi }\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {1}{1-\dfrac {e}{\pi }}=\dfrac {\pi }{\pi -e}$
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