Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.4 - Comparison Tests - Exercises 10.4 - Page 591: 41

Answer

Divergent

Work Step by Step

Re-write the given series as follows: $\Sigma_{n=1}^\infty \dfrac{2^n-n}{n 2^n}= \Sigma_{n=1}^\infty\dfrac{2^{n}}{n2^n} -\Sigma_{n=1}^\infty\dfrac{n}{n2^n}$ $\implies \Sigma_{n=1}^\infty \dfrac{2^n-n}{n 2^n}=\Sigma_{n=1}^\infty\dfrac{1}{n} -\Sigma_{n=1}^\infty (\dfrac{1}{2})^n$ Here, the first part of the series, that is, $\Sigma_{n=1}^\infty\dfrac{1}{n}$ shows a divergent p-series with $p=1$ and the second part of the series, that is, $\Sigma_{n=1}^\infty (\dfrac{1}{2})^n$ shows a geometric convergent series and has common ratio $r =\dfrac{1}{2} \lt 1$. Therefore, the sum of the series is a divergent series.
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