Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 727: 45

Answer

$b\lt \frac{1}{e}$ (with $b\gt 0$)

Work Step by Step

$\Sigma_{n=1}^{\infty}b^{ln(n)}=\Sigma_{n=1}^{\infty}n^{lnb}=\Sigma_{n=1}^{\infty}\frac{1}{n^{-lnb}}$ This is a p-series with $p=-lnb$ So if $p=-lnb\gt 1$, then the series will converge. $$-lnb\gt 1$$$$lnb\lt -1$$$$b\lt e^{-1}$$$$b\lt \frac{1}{e}$$ Hence, $b\lt \frac{1}{e}$ (with $b\gt 0$).
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