University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.4 - Comparison Tests - Exercises - Page 509: 6



Work Step by Step

Consider $a_n=\dfrac{1}{n3^{n}}$ When we increase the numerator, the value of the fraction will always increase and when we decrease the denominator, the value of the fraction will always increase. Thus, we have $a_n \leq \dfrac{1}{3^{n}}$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{3^n}$ is a convergent geometric series with common ratio $r=\dfrac{1}{3} \gt 1$. Thus the series converges by the comparison test.
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