Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.3 Convergence of Series with Positive Terms - Preliminary Questions - Page 555: 5


No, Ralph is not on the right track.

Work Step by Step

No, Ralph is not right because the series $\Sigma_{n=1}^\infty \frac{1}{n}$ diverges and the series $\Sigma_{n=1}^\infty \frac{e^{-n}}{n}=\Sigma_{n=1}^\infty \frac{1}{ne^n}$ is smaller than $\Sigma_{n=1}^\infty \frac{1}{n}$. The fact that the larger series diverges does not mean that the smaller one diverges.
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