University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.5 - Absolute Convergence; The Ratio and Root Tests - Exercises - Page 515: 22


Diverges and $\lim\limits_{n \to \infty} a_n=\dfrac{1}{e^2}$

Work Step by Step

Consider $a_n=(\dfrac{n-2}{n})^n$ Now, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (\dfrac{n-2}{n})^n$ Thus, we have $\lim\limits_{n \to \infty}(\dfrac{n}{n}-\dfrac{2}{n})=\lim\limits_{n \to \infty} (1-\dfrac{2}{n})$ So, $\lim\limits_{n \to \infty} a_n=e^{-2}=\dfrac{1}{e^2}$ Thus, $\lim\limits_{n \to \infty} a_n =0$ for the series to converge, but we can see that $\lim\limits_{n \to \infty} a_n \ne 0$ Hence, The Series Diverges and $\lim\limits_{n \to \infty} a_n=\dfrac{1}{e^2}$
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.