Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1050: 88


$ \approx 2.7183$

Work Step by Step

We know that $a_n=1+(\dfrac{1}{n})^n$ Here, we have $a_{10}=1+(\dfrac{1}{10})^{10}=2.5937$ $a_{100}=1+(\dfrac{1}{100})^{100}=2.7048$ $a_{1000}=1+(\dfrac{1}{1000})^{1000}=2.7169$ $a_{10000}=1+(\dfrac{1}{10000})^{10000}=2.7181$ ..... and so on Hence, we can see that as the value of n increases, the value approaches to $2.7183$
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.