Problem 42 - Calculus: Early Transcendentals

Question

Determine whether the sequence converges or diverges. If it converges, find the limit. \( a_n = \ln (n + 1) - \ln n \)

Step-by-Step Solution

Answer

The sequence converges to 0.

Step 1: Simplify the Sequence

Simplify the sequence expression \( a_n = \ln(n+1) - \ln n \) using the properties of logarithms. Recall the property that \( \ln(a) - \ln(b) = \ln\left( \frac{a}{b} \right) \). Applying this, we have:\[a_n = \ln \left( \frac{n+1}{n} \right).\]

Step 2: Simplify the Fraction

Simplify the fraction inside the logarithm:\[\frac{n+1}{n} = \frac{n}{n} + \frac{1}{n} = 1 + \frac{1}{n}.\]

Step 3: Analyze the Limit

Analyze the behavior of the simplified expression as \( n \to \infty \). Note that as \( n \) grows larger, \( \frac{1}{n} \) approaches 0, making the expression inside the logarithm approach 1. Hence, we have:\[a_n = \ln \left( 1 + \frac{1}{n} \right) \to \ln(1) = 0\]as \( n \to \infty \).

Step 4: Conclusion

Since \( a_n \to 0 \) as \( n \to \infty \), the sequence converges. The limit of the sequence \( a_n \) is 0.

Key Concepts

Properties of Logarithms

Logarithms are a mathematical tool used to simplify complex multiplicative relationships into additive ones. One important property of logarithms is the difference rule: \( \ln(a) - \ln(b) = \ln\left( \frac{a}{b} \right) \). This property is particularly useful when you need to simplify expressions involving logarithms, as seen in the exercise provided. By applying the difference rule, the sequence \( a_n = \ln(n+1) - \ln n \) simplifies to \( \ln\left( \frac{n+1}{n} \right) \). This step condenses the expression into a simpler form that is easier to analyze and evaluate, especially when determining convergence or divergence of a sequence.
Another useful property is the power rule: \( \ln(a^b) = b \cdot \ln(a) \), which allows for powers to be brought down as coefficients. Understanding these and other properties of logarithms is key to tackling more complex calculus problems.

Simplifying Fractions

Simplifying fractions is a fundamental technique in algebra and calculus that can make expressions easier to work with. In the problem given, we started with a fraction inside the logarithm: \( \frac{n+1}{n} \). To simplify it, we separate the terms: \( \frac{n+1}{n} = \frac{n}{n} + \frac{1}{n} \). This results in the expression: \( 1 + \frac{1}{n} \).

• \( \frac{n}{n} \) simplifies to 1.
• \( \frac{1}{n} \) remains as is because it tends towards zero as \( n \) increases.

The simplification process is not just arithmetic; it reveals the structure needed to understand the limit. Simplifying like this helps in exploring the behavior of expressions as variables tend to specific values or infinity.

Limits in Calculus

In calculus, limits help determine the behavior of a function or sequence as it approaches a particular point or infinity. The limit concept is essential for understanding convergence, which determines if a sequence settles down to a particular value.
In our exercise, once the fraction within the logarithm was simplified to \( 1 + \frac{1}{n} \), we analyzed its behavior as \( n \to \infty \). Intuitively, as \( n \) increases, the term \( \frac{1}{n} \) gets smaller and closer to zero. Thus, the expression inside the logarithm approaches 1, giving us \( \ln(1) \), which equals zero.
This concept is central to calculus and sequence convergence. Knowing how to correctly find and interpret the limits of expressions is crucial in answering convergence questions and serves as the foundational skill for more advanced calculus topics.