Problem 37 - Single Variable Calculus

Question

Euler's Constant a. Show that $$ \ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} $$ and therefore, \(0<\ln (n+1)-\ln n \leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n\) Hence, deduce that the sequence \(\left\\{a_{n}\right\\}\) defined by $$ a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n $$ is bounded below. Hint: Use Inequality (2), page 754, with \(f(x)=1 / x\). b. Show that $$ \frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} d x=\ln (n+1)-\ln n $$ and use this result to show that the sequence \(\left\\{a_{n}\right\\}\) defined in part (a) is decreasing. Hint: Draw a figure similar to Figure 1 . c. Use the Monotone Convergence Theorem to show that \(\left\\{a_{n}\right\\}\) is convergent. Note: The number $$ \gamma=\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n\right) $$ whose value is \(0.5772 \ldots\), is called Euler's constant.

Step-by-Step Solution

Answer

Thus, the sequence \( \left\{a_{n}\right\} \) defined by \[ a_n=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \] is bounded below since for all \(n\), we have \(a_n \geq 0\).

Step 1: Choose the integration technique

Use substitution, parts, partial fractions, or direct integration.

Step 2: Evaluate

The integral evaluates to Thus, the sequence \( \left\{a_{n}\right\} \) defined by \[ a_n=1+\frac{1}{2}+\cdots+\frac{1}{n}-\l.

Key Concepts

Inequalities in Calculus

In calculus, inequalities are a powerful tool for establishing boundaries and comparisons between values. They are vital in analyzing functions and their behavior. For instance, in this exercise, we use inequalities to compare the natural logarithm function with a sum of fractions. This involves integrating a function over a specific interval and comparing it to a discrete sum.

We aim to show:

• \( \ln(n+1) \leq 1 + \frac{1}{2} + \cdots + \frac{1}{n} \)
• \( 0 < \ln(n+1) - \ln(n) \leq 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \ln(n) \)

These inequalities reassure us that as \(n\) grows larger, the sum of the series \(1 + \frac{1}{2} + \cdots + \frac{1}{n}\) provides an upper boundary for \(\ln(n+1)\). This boundary is crucial in understanding the convergence properties of sequences as used in estimating Euler's constant.

Integral Approximation

Integral approximation involves estimating the value of a definite integral by comparing it to something simpler, such as a sum. In this exercise, to approximate \(\ln(n+1) - \ln(n)\), we compare the integral from \(n\) to \(n+1\) of the function \(f(x) = \frac{1}{x}\) with \(\frac{1}{n+1}\):

• \( \frac{1}{n+1} < \int_{n}^{n+1} \frac{1}{x} dx = \ln(n+1) - \ln(n) \)

This approximation helps us determine how the newly added terms in a series shift the value of the natural logarithm. By understanding that the area under the curve \(\frac{1}{x}\) from \(n\) to \(n+1\) lies between fractions, we maintain a clear gauge on how the entire sequence behaves as \(n\) increases.

Monotone Convergence Theorem

The Monotone Convergence Theorem is a fundamental concept in calculus and real analysis. It posits that if a sequence is both monotonic and bounded, then it is convergent. This theorem is particularly applicable in part (c) of our exercise.

Using the theorem, we explore the sequence \(a_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \ln(n)\) and show it is monotonic and bounded:

• The sequence decreases because each additional term added in the harmonic sum is less pronounced than the increase in \(\ln(n)\).
• It is bounded below, preventing it from decreasing indefinitely.

Thus, by the Monotone Convergence Theorem, \(a_n\) converges to a limit, which is Euler's constant \(\gamma\), as \(n\) approaches infinity.

Sequence Convergence

Sequence convergence is a core idea in calculus, essential for understanding limits and behavior at infinity. A sequence converges when its terms approach a specific value as the sequence progresses.

In this exercise, the sequence \(a_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \ln(n)\) was shown to be bounded and decreasing, paving the way for convergence. Ultimately, this sequence converges to Euler's constant \(\gamma\), known to be approximately 0.5772.

• Theorems like the Monotone Convergence Theorem confirm convergence once the properties of the sequence are confirmed.
• By investigating component terms in the sequence, we assess their contribution toward the eventual limit.

Recognizing convergence through sequences helps us better model complex phenomena in mathematics, often revealing underpinned constants like Euler's, pivotal in number theory and analysis.