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: 37


$\lim\limits_{n \to \infty} a_n=\dfrac{1}{2}$ and {$a_n$} is convergent

Work Step by Step

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