/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 In Chapter \(8,\) we will encoun... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 75

In Chapter \(8,\) we will encounter the harmonic \(\operatorname{sum} 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} .\) Use a left Riemann sum to approximate \(\int_{1}^{n+1} \frac{d x}{x}(\) with unit spacing between the grid points) to show that \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}>\ln (n+1) .\) Use this fact to conclude that \(\lim _{n \rightarrow \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\) does not exist.

Short Answer

Expert verified
Question: Show that the sum of the harmonic sequence \(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\) is greater than the natural logarithm of \((n+1)\), and prove that its limit as \(n\) approaches infinity does not exist. Answer: We showed that the harmonic sum \(S_n\) is greater than the integral \(I_n = \int_{1}^{n+1} \frac{dx}{x}\) by using a left Riemann sum to approximate \(I_n\). Recognizing that the left Riemann sum was exactly the same as the harmonic sum \(S_n\), we concluded that \(S_n > I_n\). Since \(I_n\) evaluates to the natural logarithm of \(n+1\), we proved that the harmonic sum is greater than the natural logarithm. To show the limit does not exist, we noted that the limit of the natural logarithm as \(n\) approaches infinity is infinity, and since the harmonic sum is always greater, its limit as \(n\) approaches infinity does not exist, as well.

Step by step solution

01

Define the given harmonic sum

We have the harmonic sum: \(S_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\).
02

Define the integral to be approximated

We need to approximate the integral: \(I_n = \int_{1}^{n+1} \frac{dx}{x}\).
03

Approximate the integral using left Riemann sum

We will use a left Riemann sum to approximate \(I_n\). The left Riemann sum of a function is the sum of the areas of rectangles under the curve, where the height of each rectangle is the value of the function at the left endpoint of the interval, and the width of each rectangle is unit spacing. For this problem, the left Riemann sum is: \(R_n = \sum_{k=1}^{n} \frac{1}{k} (1)\), where the interval width is \(1\).
04

Show that the harmonic sum is greater than the left Riemann sum

Notice that for each term in the harmonic sum: \(S_n = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\), the corresponding term in the left Riemann sum is: \(R_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\), which is exactly the same as the harmonic sum \(S_n\). Since \(I_n\) can be approximated by the left Riemann sum \(R_n\) and \(S_n = R_n\), the harmonic sum is greater than the integral. \(S_n = R_n > I_n\).
05

Show that the harmonic sum is greater than the natural logarithm

Since \(I_n = \int_{1}^{n+1} \frac{dx}{x}\) and we have shown that \(S_n > I_n\), then it is also true that \(S_n > \ln(n+1)\), since the integral evaluates to: \(I_n = \left[\ln(x)\right]_1^{n+1} = \ln(n+1) - \ln(1) = \ln(n+1)\).
06

Conclude the limit does not exist

To show that the limit of the harmonic sum as \(n\) approaches infinity does not exist, note that since \(S_n > \ln(n+1)\) for all \(n\), and the limit of \(\ln(n+1)\) as \(n\) approaches infinity is infinity, it follows that the limit of \(S_n\) as \(n\) approaches infinity also does not exist. In other words: \(\lim _{n \rightarrow \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\) does not exist.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Riemann Sum
The Riemann Sum is a method for approximating the total area under a curve. It's commonly used in calculus to estimate integrals, especially when the exact integral is difficult to solve directly. In this problem, we are tasked with approximating the integral of \( \frac{1}{x} \) from \(1\) to \(n+1\) using a left Riemann sum. This approximation considers dividing the area into rectangles, where each rectangle's height is determined by the function's value at the left endpoint of each subinterval. The width of each rectangle is the distance between consecutive points, which in this case is 1. Therefore, the sum \(R_n = \sum_{k=1}^{n} \frac{1}{k} \times 1\) gives us the left Riemann sum. This approach helps us compare the harmonic sum to the definite integral of a natural logarithm expression, providing valuable insight into the behavior of the harmonic series.
Convergence
The concept of convergence is crucial when dealing with infinite series, like the harmonic series \(1 + \frac{1}{2} + \frac{1}{3} + \cdots\). A series converges if the sum of its terms approaches a specific finite number as more terms are added. On the other hand, the harmonic series is a classic example of a divergent series. This divergence is demonstrated by comparing the harmonic sum \(S_n\) with \(\ln(n+1)\). We found that \(S_n > \ln(n+1)\), and since \(\ln(n+1)\) approaches infinity as \(n\) increases, so does \(S_n\). Consequently, the harmonic series does not converge because it does not settle towards a finite limit, unlike some other series that can converge to a defined value.
Definite Integral
A definite integral is a fundamental concept in calculus. It calculates the accumulated quantity, such as area under a curve, within a specified limit. In our exercise, we are evaluating the definite integral \(\int_{1}^{n+1} \frac{dx}{x}\), which gives us \(\ln(n+1)\). This outcome is important as it provides a reference point when comparing with the harmonic sum \(S_n\). To evaluate this integral, we find the antiderivative of \(\frac{1}{x}\), which is \(\ln(x)\). Then, we substitute the limits to determine \(\ln(n+1) - \ln(1)\), simplified to \(\ln(n+1)\). This result is used to show that the harmonic sum exceeds this natural logarithm value, leading to the conclusion about the divergence of the harmonic series. Understanding definite integrals is essential for connecting the dots between continuous and discrete mathematical concepts in this problem.

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