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 488: 39

Answer

$\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent.

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{(-1)^{n+1}}{2n-1}$ This implies $\lim\limits_{n \to \infty} \dfrac{\dfrac{(-1)^{n+1}}{2n-1}}{\dfrac{2n}{n}-\dfrac{1}{n}}=\dfrac{0}{2}=0$ Thus, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent.
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