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 - Practice Exercises - Page 636: 2

Answer

sequence converges to $0$.

Work Step by Step

As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists. Consider $a_n=\dfrac{1-(-1)^n}{\sqrt n}$ Apply limits to both sides. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}[\dfrac{1-(-1)^n}{\sqrt n}]$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{2}{\sqrt n}$ $\lim\limits_{n \to \infty}a_n=\dfrac{2}{\sqrt \infty}$ $\lim\limits_{n \to \infty}a_n=\dfrac{2}{ \infty}$ $\lim\limits_{n \to \infty}a_n=0$ Hence, the sequence converges to $0$.
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.