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: 41



Work Step by Step

Since, we have $\Sigma_{n=1}^\infty \dfrac{2^n-n}{n 2^n}$ and $\Sigma_{n=1}^\infty \dfrac{2^n-n}{n 2^n}= \Sigma_{n=1}^\infty\dfrac{2^{n}}{n2^n} -\Sigma_{n=1}^\infty\dfrac{n}{n2^n}$ or, $=\Sigma_{n=1}^\infty\dfrac{1}{n} -\Sigma_{n=1}^\infty (\dfrac{1}{2})^n$ Thus, the series $\Sigma_{n=1}^\infty\dfrac{1}{n}$ is a divergent p-series with $p=1$ and $\Sigma_{n=1}^\infty (\dfrac{1}{2})^n$ is a convergent series because it is a geometric convergent series with common ratio $r =\dfrac{1}{2} \lt 1$ Hence, the sum of the two series diverges.
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