Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 9 - Infinite Series - 9.6 Exercises - Page 634: 55

Answer

Diverges

Work Step by Step

Here, we have the given series $\Sigma_{n=1}^{\infty} \dfrac{5 n}{2n-1}$ Next, we will use the nth divergence test for the given series to check whether it is convergent or divergent. $\lim\limits_{n \to \infty} \dfrac{5n }{2n-1}=\lim\limits_{n \to \infty} \dfrac{5}{2-1/n}=\dfrac{5}{2+0}=\dfrac{5}{2} \ne 0$ Hence, we can conclude that the given series diverges by the nth divergence test .
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