Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.3 The Integral Test and Estimates of Sums - 11.3 Exercises - Page 767: 45


$b\lt \frac{1}{e}$ (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 $b\gt 0$).
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.