Problem 75 - Calculus Early Transcendentals

Question

The harmonic sum is \(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n} .\) Use a right Riemann sum to approximate \(\int_{1}^{n} \frac{d x}{x}(\) 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.

Step-by-Step Solution

Answer

Question: Show that the harmonic sum is greater than the natural logarithm of (n+1), and use this to argue that the limit of the harmonic sum as n approaches infinity does not exist. Answer: We have shown that the harmonic sum is greater than the logarithm function: \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} > \ln(n+1)\). As n approaches infinity, the logarithm function grows to infinity. Since the harmonic sum is always greater than the logarithm function, it will also grow to infinity as n approaches infinity. Therefore, the limit of the harmonic sum as n approaches infinity does not exist.

Step 1: Understanding Riemann sums

A Riemann sum is a method to approximate the definite integral of a function using a finite sum. In this case, we will use a right Riemann sum, which means that we will compute the area of rectangles with heights determined by the function values at the right endpoints of each subinterval.

Step 2: Setting up the right Riemann sum

We want to approximate the integral of \(\frac{1}{x}\) from \(1\) to \(n\) using unit spacing. To set this up, first divide the interval \([1, n]\) into \(n-1\) subintervals of equal length, which in this case is 1. The right endpoints of these subintervals are \(x_i = 2, 3, \dots, n\). Now apply the right Riemann sum formula: \(\int_{1}^{n} \frac{1}{x} dx \approx \sum_{i=1}^{n-1} \frac{1}{x_i} \Delta x_i = \sum_{i=1}^{n-1} \frac{1}{i+1}\)

Step 3: Comparing the right Riemann sum to the harmonic sum

Now, we want to compare the right Riemann sum, which is an approximation of the integral, to the harmonic sum. Notice that: \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} = \sum_{i=1}^n \frac{1}{i} > \sum_{i=1}^{n-1} \frac{1}{i+1}\) This is because we have more positive terms in the harmonic sum than in the right Riemann sum, and each term in the right Riemann sum can be paired with a corresponding term in the harmonic sum.

Step 4: Integrating the function and comparing the result

Now we will integrate the function and compare the result to our inequality: \(\int_{1}^{n} \frac{1}{x} dx = [\ln|x|]_{1}^{n} = \ln(n) - \ln(1) = \ln(n)\). Recall that the right Riemann sum is an approximation of this integral: \(\sum_{i=1}^{n-1} \frac{1}{i+1} \approx \int_{1}^{n} \frac{1}{x} dx = \ln(n)\). Since the harmonic sum is greater than the right Riemann sum and the right Riemann sum approximates the integral, we can conclude that: \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} > \ln(n) > \ln(n+1)-\ln(1)=\ln(n+1)\).

Step 5: Concluding that the limit does not exist

We have shown that the harmonic sum is greater than the logarithm function: \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} > \ln(n+1)\). As \(n\) approaches infinity, the logarithm function grows to infinity. Since the harmonic sum is always greater than the logarithm function, it will also grow to infinity as \(n\) approaches infinity. Therefore, we can conclude that: \(\lim_{n\rightarrow \infty} \left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\) does not exist.

Key Concepts

Harmonic Sum

The harmonic sum is a fascinating concept in mathematics, represented as \(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\). It is the sum of the reciprocals of the first \(n\) natural numbers. This sequence adds each term as the reciprocal of an increasing integer.
This sum grows slowly but steadily, becoming larger as \(n\) increases. However, unlike other sequences that converge to a particular value, the harmonic sum does not have a limit as \(n\) approaches infinity. It continues to grow without bound.

• The harmonic sum is a key player in analyzing the behavior of sequences and series in math.
• It appears in various algorithms and calculations, including computer science and number theory.

Understanding this sum helps in grasping more profound mathematical patterns and behaviors.

Definite Integral

A definite integral is a fundamental concept in calculus, representing the accumulated area under a curve within a specific interval. In mathematical terms, the definite integral from \(a\) to \(b\) of a function \(f(x)\) is given by \(\int_{a}^{b} f(x) \, dx\). This integral calculates the total change represented by the function over the interval.
In the exercise, the definite integral \(\int_{1}^{n} \frac{1}{x} \, dx\) is compared with the harmonic sum. Here’s why it's important:

• It offers an exact mathematical way to understand the growth of the function \(\frac{1}{x}\).
• It is used to approximate the accumulation of values over a range, unlike a finite sum.

Interpreting definite integrals allows us to grasp concepts of accumulation and change, which are vital in many fields such as physics, engineering, and economics.

Logarithmic Function

Logarithmic functions invert the process of exponentiation and are represented as \(\ln(x)\) for natural logarithms. In the context of the harmonic sum and its comparison to integrals, logarithmic functions help illustrate growth behavior.
The integral \(\int_{1}^{n} \frac{1}{x} \, dx\) results in \(\ln(n)\), which then provides a method to compare it to the harmonic sum:

• \(\ln(x)\) grows, albeit slower than linear functions, allowing for comparison with the harmonic sum.
• By establishing \(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} > \ln(n+1)\), we observe that the harmonic sum grows faster than the logarithm.

Understanding logarithmic functions reveals many complex behaviors in mathematics and nature, showing how quantities grow and scale over time.

Limit of a Sequence

The concept of a limit for a sequence is a cornerstone of mathematical analysis. It describes the behavior of a sequence as its index approaches infinity. For the harmonic sum \(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\), we determine whether a limit exists or not.
Our task was to show that

• The harmonic sum exceeds the logarithmic growth \(\ln(n+1)\).
• Since \(\ln(x)\) tends towards infinity as \(x\) increases, the harmonic sum does too.

This absence of a limit means the harmonic sum diverges, unlike convergent sequences that settle towards a fixed value. Understanding whether a sequence converges or diverges helps in analyzing complex systems and predicting their behavior over time.