Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - Chapter Review Exercises - Page 592: 80


The series $\sum_{n=1}^{\infty} ne^{-0.02n}$ converges.

Work Step by Step

We apply the ratio test as follows, $a_n=ne^{-0.02n}$ $$ \rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty}\frac{(n+1)e^{-0.02}}{n}\\ =\lim _{n \rightarrow \infty}(1+(1/n))e^{-0.02}=e^{-0.02}=0.98\lt 1 $$ Hence the series $\sum_{n=1}^{\infty} ne^{-0.02n}$ converges.
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