Explanation Given : The sequence b n = n ! n n Calculation : Step 1: b n = n ! n n can be rewritten as b n = ( n ! ) 1 n n Step 2: Taking natural logarithms on both sides, we get ln b n = ln ( n ! ) 1 n − ln n = 1 n ln ( n ! ) − ln n = ln ( n ! ) − n ln n n = 1 n [ ln ( n ! ) − ln n n ] = 1 n ln n ! n n = 1 n ln ( 1 n ... To determine (b) The mathematical proof of the statement stating that ln ( b n ) converges to ∫ 0 1 ln x d x if b n = n ! n n and that b n → e − 1

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 11.1, Problem 90E

To determine

(a)

The mathematical proof of the relation $\ln (b_{n}) = \frac{1}{n} \sum_{k = 1}^{n} \ln \frac{k}{n}$ if $b_{n} = \frac{\sqrt[n]{n!}}{n}$

To determine

(b)

The mathematical proof of the statement stating that $\ln (b_{n})$ converges to $\int_{0}^{1} \ln x d x$ if $b_{n} = \frac{\sqrt[n]{n!}}{n}$ and that $b_{n} \to e^{- 1}$

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Chapter 11.1, Problem 89E

Chapter 11.1, Problem 91E

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Chapter 11 Solutions

CALCULUS (CLOTH)

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

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(a) The mathematical proof of the relation ln ( b n ) = 1 n ∑ k = 1 n ln k n if b n = n ! n n

Solution Summary: The author explains how the mathematical proof of the relation ln (b n) is mathematically proved.

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(a)

The mathematical proof of the relation $\ln (b_{n}) = \frac{1}{n} \sum_{k = 1}^{n} \ln \frac{k}{n}$ if $b_{n} = \frac{\sqrt[n]{n!}}{n}$

(b)

The mathematical proof of the statement stating that $\ln (b_{n})$ converges to $\int_{0}^{1} \ln x d x$ if $b_{n} = \frac{\sqrt[n]{n!}}{n}$ and that $b_{n} \to e^{- 1}$

Want to see the full answer?

Chapter 11 Solutions

(a) The mathematical proof of the relation ln ( b n ) = 1 n ∑ k = 1 n ln k n if b n = n ! n n

Solution Summary: The author explains how the mathematical proof of the relation ln (b n) is mathematically proved.

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(a) The mathematical proof of the relation ln(bn)=1n∑k=1nlnkn if bn=n!nn

(b) The mathematical proof of the statement stating that ln(bn) converges to ∫01lnxdx if bn=n!nn and that bn→e−1

Want to see the full answer?

Chapter 11 Solutions

(a)

The mathematical proof of the relation $\ln (b_{n}) = \frac{1}{n} \sum_{k = 1}^{n} \ln \frac{k}{n}$ if $b_{n} = \frac{\sqrt[n]{n!}}{n}$

(b)

The mathematical proof of the statement stating that $\ln (b_{n})$ converges to $\int_{0}^{1} \ln x d x$ if $b_{n} = \frac{\sqrt[n]{n!}}{n}$ and that $b_{n} \to e^{- 1}$