University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 68

Answer

Divergent

Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ The given series has first term, $a=1$ and common ratio $r =|\dfrac{e^\pi}{\pi^e}|=\dfrac{23.141}{22.459}$ Hence, the given series diverges because $|r|=|\dfrac{23.141}{22.459}| \gt 1$
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