×
Log in to StudySoup
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.7 - Problem 86ae
Join StudySoup for FREE
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.7 - Problem 86ae

Already have an account? Login here
×
Reset your password

Riemann sums to integrals Show that in the following

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 86AE Chapter 7.7

Calculus: Early Transcendentals | 1st Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

4 5 1 350 Reviews
29
3
Problem 86AE

Riemann sums to integrals Show that \(L=\lim _{n \rightarrow \infty}\left(\frac{1}{n} \ln n !-\ln n\right)=-1\) in the following steps.

a. Note that \(n !=n(n-1)(n-2) \cdots 1\) and use ln (ab)=ln a + ln b to show that \(L=\lim _{n \rightarrow \infty}\left[\left(\frac{1}{n} \sum_{k=1}^{n} \ln k\right)-\ln n\right]=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(\frac{k}{n}\right)\)

b. Identify the limit of this sum as a Riemann sum for \(\int_{0}^{1} \ln \ x \ d x\). Integrate this improper integral by parts and reach the desired conclusion.

Step-by-Step Solution:
Step 1 of 3

Solution:-

Step1

To find

Show that  in the following steps.

a. Note that n! = n(n − 1)(n − 2) ··· 1 and use ln (ab)=ln a + ln b to show that

b. Identify the limit of this sum as a Riemann sum for . Integrate this improper integral by parts and reach the desired conclusion.

Step2

a. Note that n! = n(n − 1)(n − 2) ··· 1 and use ln (ab)=ln a + ln b to show that

L=

  =

 =

L=

=

=

=

=

Hence,

L=

Step3

b. Identify the limit of this sum as a Riemann sum for . Integrate this improper integral by parts and reach the desired conclusion.

=

                       =

                       =

                     =

                    = -1

Hence

L===-1

Step 2 of 3

Chapter 7.7, Problem 86AE is Solved
Step 3 of 3

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

The full step-by-step solution to problem: 86AE from chapter: 7.7 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. The answer to “?Riemann sums to integrals Show that \(L=\lim _{n \rightarrow \infty}\left(\frac{1}{n} \ln n !-\ln n\right)=-1\) in the following steps.a. Note that \(n !=n(n-1)(n-2) \cdots 1\) and use ln (ab)=ln a + ln b to show that \(L=\lim _{n \rightarrow \infty}\left[\left(\frac{1}{n} \sum_{k=1}^{n} \ln k\right)-\ln n\right]=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(\frac{k}{n}\right)\)b. Identify the limit of this sum as a Riemann sum for \(\int_{0}^{1} \ln \ x \ d x\). Integrate this improper integral by parts and reach the desired conclusion.” is broken down into a number of easy to follow steps, and 79 words. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. Since the solution to 86AE from 7.7 chapter was answered, more than 464 students have viewed the full step-by-step answer. This full solution covers the following key subjects: riemann, sum, show, Note, integral. This expansive textbook survival guide covers 112 chapters, and 7700 solutions. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567.

Other solutions

Discover and learn what students are asking











Statistics: Informed Decisions Using Data : Comparing Three or More Means (One-Way Analysis of Variance)
?In Problems 7 and 8, fill in the ANOVA table.

Statistics: Informed Decisions Using Data : Testing the Significance of the Least-Squares Regression Model
?In the least-squares regression model, yi = b1xi + b0 + ei, ei is a random error term with mean ______________ and standard deviation aet = __________

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Riemann sums to integrals Show that in the following