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: 46


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

Work Step by Step

Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{\sin^2 n}{2^n}$ Since, we know that $-1 \leq \sin n \leq 1$ , thus apply Sandwich Theorem, which states that $\lim\limits_{n \to 0} \dfrac{1}{n}=0$ and $-1 \leq \sin^2 n \leq 1$ So, $ \lim\limits_{n \to \infty}\dfrac{-1}{2^n}=0$ This implies that, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{\sin^2 n}{2^n}=0$ Thus, $\lim\limits_{n \to \infty} a_n=0 $ 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.