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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.7 - Problem 86ae
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.7 - Problem 86ae

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# Riemann sums to integrals Show that in the following

ISBN: 9780321570567 2

## Solution for problem 86AE Chapter 7.7

Calculus: Early Transcendentals | 1st Edition

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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

Step 3 of 3

##### 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.

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Riemann sums to integrals Show that in the following