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 570: 37


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

Work Step by Step

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