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

Answer

$\lim\limits_{n \to \infty} a_n=\dfrac{1}{2}$ 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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.