University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 489: 117

Answer

$\{a_{n}\}$ is bounded and monotonic$\Rightarrow$ it is convergent.

Work Step by Step

$\{a_{n}\}$ is bounded above by 1, because in $\displaystyle \frac{2^{n}-1}{2^{n}},$ the numerator is smaller than the denominator (and both are positive). $\{a_{n}\}$ is bounded below by 0. If we can show that $\{a_{n}\}$ is monotonic, we can deduce by Th.6 that $\{a_{n}\}$ is convergent. \begin{align*} a_{n+1}-a_{n}&=\displaystyle \frac{2^{n+1}-1}{2^{n+1}}-\frac{2^{n}-1}{2^{n}} \\ & =\displaystyle \frac{2^{n+1}-1}{2^{n+1}}-\frac{2(2^{n}-1)}{2(2^{n})} \\ & =\displaystyle \frac{2^{n+1}-1-(2^{n+1}-2)}{2^{n+1}}\\ & =\displaystyle \frac{1}{2^{n+1}}\geq 0\\ a_{n+1}&\geq a_{n} \\ \end{align*} It is monotonic. Thus, it is convergent.
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