Problem 63 - Thomas Calculus

Question

Euler's constant Graphs like those in Figure 10.12 suggest that as \(n\) increases there is little change in the difference between the sum $$ 1+\frac{1}{2}+\cdots+\frac{1}{n} $$ and the integral $$\ln n=\int_{1}^{n} \frac{1}{x} d x$$ To explore this idea, carry out the following steps. $$ \begin{array}{c}{\text { a. By taking } f(x)=1 / x \text { in the proof of Theorem } 9, \text { show that }} \\ {\ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} \leq 1+\ln n} \\ {\text { or }} \\ {0<\ln (n+1)-\ln n \leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \leq 1} \\ {\text { Thus, the sequence }} \\ {a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n} \\\ {\text { is bounded from below and from above. }}\end{array} $$ $$ \begin{array}{l}{\text { b. Show that }} \\ {\frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} d x=\ln (n+1)-\ln n} \\ {\text { and use this result to show that the sequence }\left\\{a_{n}\right\\} \text { in part (a) }} \\ {\text { is decreasing. }} \\ {\text { since a decreasing sequence that is bounded from below }} \\ {\text { converges, the numbers } a_{n} \text { defined in pounded from below }} \\ {\text { converges, the numbers } a_{n} \text { defined in part (a) converge: }}\end{array} $$ $$ \begin{array}{c}{1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \rightarrow \gamma} \\\ {\text { The number } \gamma, \text { whose value is } 0.5772 \ldots, \text { is called Euler's }} \\ {\text { constant. }}\end{array} $$

Step-by-Step Solution

Answer

The sequence \( a_n \) converges to Euler's constant, \( \gamma \approx 0.5772 \).

Step 1: Prove the Upper Bound

We need to show that \(1 + \frac{1}{2} + \cdots + \frac{1}{n} \leq 1 + \ln n\). This is achieved using the integral approximation, where \( \int_{1}^{n} \frac{1}{x} \, dx = \ln n \). Since the function \( f(x) = \frac{1}{x} \) is decreasing, the right endpoint approximation method implies that the sum of the series is at most \( 1 + \ln n \).

Step 2: Prove the Lower Bound

We need to show that \( \ln(n+1) \leq 1 + \frac{1}{2} + \cdots + \frac{1}{n} \). By using the left endpoint approximation for the integral of a decreasing function, we arrive at the conclusion that: \( \int_{1}^{n+1} \frac{1}{x} \, dx \leq 1 + \frac{1}{2} + \cdots + \frac{1}{n}\). Therefore, \(\ln(n+1) \leq 1 + \frac{1}{2} + \cdots + \frac{1}{n} \).

Step 3: Define and Bound Sequence

From Steps 1 and 2, we get: \(0 < \ln(n+1) - \ln n \leq 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \ln n \leq 1\). This shows the sequence \(a_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \ln n\) is bounded from both below (positive lower bound) and above (by 1).

Step 4: Show Integral Inequality

We need to demonstrate that \( \frac{1}{n+1} < \int_{n}^{n+1} \frac{1}{x} \, dx = \ln(n+1) - \ln n \). This is due to the fact that the average value of a decreasing function \(f(x) = 1/x\) over the interval \([n, n+1]\) is greater than the end point of the interval value \(1/(n+1)\).

Step 5: Prove Sequence is Decreasing

Since \( \frac{1}{n+1} < \ln(n+1) - \ln n \), it follows that \( 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \ln n > 1 + \frac{1}{2} + \cdots + \frac{1}{n+1} - \ln(n+1) \). Therefore, the sequence \(a_n\) is decreasing.

Step 6: Convergence to Euler's Constant

A decreasing sequence that is bounded below must converge. Therefore, \(a_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \ln n \) converges to a number \(\gamma\), known as Euler's constant, with approximate value 0.5772.

Key Concepts

Harmonic Series

The harmonic series is a well-known mathematical series. It consists of the sum of the reciprocals of the natural numbers:

• First term: 1
• Second term: \( \frac{1}{2} \)
• Third term: \( \frac{1}{3} \)
• ...and so forth up to \( \frac{1}{n} \).

The harmonic series can be represented as \[1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\].
As \( n \) grows larger, this series increases steadily without approaching a fixed limit.
However, its growth rate is slow and logarithmic in nature. Thus, the harmonic series is divergent, meaning it does not converge to a particular value. This series often serves as a foundational concept in understanding phenomena such as the harmonic mean and various approximations in mathematics.

Logarithmic Integral

The logarithmic integral, often denoted as \( \ln n \), is a vital concept in calculus and number theory. It represents the integral of the function \( \, \frac{1}{x} \, \) from 1 to \( n \), expressed as: \[\int_{1}^{n} \frac{1}{x} \, dx = \ln n.\] This integral approximation is essential for comparing with the harmonic series to derive inequalities and bounds.
It provides insights on how slowly growing functions behave over increasing intervals.
One of the key properties is that it accurately describes the accumulation of changes over the range, making it a crucial tool in approximating various mathematical series and evaluating their growth patterns.

Decreasing Sequence

A sequence is termed decreasing when each term is less than or equal to the preceding term. In mathematical terms, a sequence \( a_n \) is decreasing if \( a_{n+1} \leq a_n \) for all integers \( n \).
In the problem discussed, the sequence \[a_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \ln n\]has been shown to be decreasing.
This is because as \( n \) increases, each additional fraction \( \frac{1}{n+1} \) becomes smaller, yet the total subtraction due to the logarithm grows.
Decreasing sequences are particularly important because, when combined with the property of being bounded from below, they must converge to a limit. Such sequences are considered stable in their respective boundaries, useful for analyzing limits and convergences in mathematical expressions.

Bounded Sequence

A sequence is said to be bounded if there exist real numbers (bounds) such that every term in the sequence lies between these bounds. Specifically,
a sequence \( a_n \) is bounded below if there is a number \( L \) such that \( a_n \geq L \) for all \( n \),
and it is bounded above if there is a number \( U \) such that \( a_n \leq U \) for all \( n \).
In the context of this exercise, the sequence \[a_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \ln n\]was proven to be bounded; it exists within the range between 0 and 1.
The notion of boundedness assures us that the sequence will not diverge and will stay contained within a specific interval, making it suitable for convergence analysis. Bounded sequences often provide grounds for applying various mathematical theorems, particularly those related to convergence and stability.