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 10.1 - Sequences - Exercises 10.1 - Page 571: 118

Answer

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

Work Step by Step

$\{a_{n}\}$ is bounded below by 0, as all terms are positive. $\{a_{n}\}$ is bounded above by 1, because in $\displaystyle \frac{2^{n}-1}{3^{n}},$ the numerator is smaller than the denominator (and both are positive). 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}{3^{n+1}}-\frac{2^{n}-1}{3^{n}} \\ & =\displaystyle \frac{2^{n+1}-1}{3^{n+1}}-\frac{3(2^{n}-1)}{3(3^{n})} \\ & =\displaystyle \frac{2^{n+1}-1-(3\cdot 2^{n+1}-3)}{3^{n+1}}\\ & =\displaystyle \frac{2-2^{n+1}}{3^{n+1}}\leq 0\\ a_{n+1}&\leq a_{n} \\ \end{align*} It is monotonic. Thus, it is convergent.
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