Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.7 - Power Series - Exercises 10.7 - Page 613: 46


Interval of Convergence : $e^{-1} \lt x \lt e$ and $\Sigma_{n=0}^{\infty} (\ln x)^n=\dfrac{1}{1-\ln x}$

Work Step by Step

Remember that the series will converge when $|f(x)|=|\ln x| \lt 1$ This implies that $-1 \lt \ln x \lt 1$ and $e^{-1} \lt x \lt e$ This means that the interval of convergence is: $e^{-1} \lt x \lt e$ Next, we will find the sum of the given geometric series $S=\dfrac{a}{1-r}=\dfrac{1}{1-\ln x}$ and $\Sigma_{n=0}^{\infty} (\ln x)^n=\dfrac{1}{1-\ln x}$
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.