Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 538: 81


The sequence $b_n=a_{n+1}$ also converges.

Work Step by Step

When $\lim\limits_{x \to \infty} f(x)$ exists, then the sequence $a_n=f(n)$ converges to the same limit, so we have: $ \lim\limits_{n \to \infty} a_n=L$ We have: $\lim\limits_{n \to \infty} (a_n -L)=0$ or, $\lim\limits_{n \to \infty} (a_n+a_{n+1} -a_{n+1} -L)=0$ or, $\lim\limits_{n \to \infty} (a_{n+1} - L) + \lim\limits_{n \to \infty} (a_n-a_{n+1})=0$ or, $\lim\limits_{n \to \infty} (a_{n+1} - L) =0$ We are given that: $a_{n+1}=b_n$ So, $\lim\limits_{n \to \infty} (b_n - L) =0 \implies \lim\limits_{n \to \infty} b_n = L$ We see that the sequence $b_n=a_{n+1}$ also converges.
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.