Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 11 - Section 11.4 - The Comparison Tests - 11.4 Exercises - Page 765: 50

Answer

$a_{n}=\frac{1}{n^{2}}$, $b_{n}=\frac{1}{n}$

Work Step by Step

For a positive convergent series we can say use $a_{n}=\frac{1}{n^{2}}$ (p-series with $p\gt 1$) For a positive divergent series we can say use $b_{n}=\frac{1}{n}$ (p-series with $p= 1)$ Then we have $\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{\frac{1}{n^{2}}}{\frac{1}{n}}$ $=\lim\limits_{n \to \infty}\frac{1}{n}$ $=0$ Hence, $a_{n}=\frac{1}{n^{2}}$, $b_{n}=\frac{1}{n}$
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