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: 24

Answer

$=\dfrac {\pi }{\pi -1}$

Work Step by Step

$1+\dfrac {1}{\pi }+\dfrac {1}{\pi ^{2}}+\dfrac {1}{\pi ^{3}}...=\dfrac {a_{1}}{1-r}=\dfrac {1}{1-\dfrac {1}{\pi }}=\dfrac {\pi }{\pi -1}$
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