Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.1 Exercises - Page 726: 89

Answer

$\lim\limits_{n\to\infty} (a_nb_n)=0$

Work Step by Step

Recall the property of the absolute value: $-|a_n| \leq a_n \leq |a_n|$ for all the values of $n$ consider $\lim\limits_{n\to\infty} |a_n|=0$ By the limit laws of a sequences, we have $\lim\limits_{n\to\infty} |a_n|=0$; $\lim\limits_{n\to\infty} -|a_n|=-\lim\limits_{n\to\infty} |a_n|=0$ By the squeeze theorem for the sequence, we have $\lim\limits_{n\to\infty} a_n=0$. Thus, If $\lim\limits_{n\to\infty} |a_n|=0$, then $\lim\limits_{n\to\infty} a_n=0$ Consider a small number $\epsilon \gt 0$ , then $a_n \to 0$ There will exist a number $N$ such that $|a_n| \lt \dfrac{\epsilon }{M}$ for the all values of $n \geq N$ $|a_nb_n|=|a_n||b_n| \lt \dfrac{\epsilon }{M} (M)=\epsilon$ Hence, it has been proved that $\lim\limits_{n\to\infty} (a_nb_n)=0$
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.