University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 84


Converges to $1$

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \sqrt [n] {n^2+n}$ Since, $ e^{\ln x}=x$ So, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \sqrt [n] {n^2+n}=\lim\limits_{n \to \infty} e^{\frac{\ln [n(n+1)]}{n}}$ or, $=e^{\lim\limits_{n \to \infty} \frac{2n+1}{n^2+n} (\frac{1/n^2}{1/n^2})}$ or, $=e^0$ Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} is Convergent and converges to $1$
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.