Solution Found!
Riemann sums to integrals Show that in the following
Chapter 7, Problem 86AE(choose chapter or problem)
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.
Questions & Answers
QUESTION:
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.
ANSWER: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.