Problem 49 - Calculus: Early Transcendentals

Question

Determine whether the sequence converges or diverges. If it converges, find the limit. \( a_n = \ln(2n^2 + 1) - \ln(n^2 + 1) \)

Step-by-Step Solution

Answer

The sequence converges to \( \ln(2) \).

Step 1: Simplify the expression

The given sequence is \( a_n = \ln(2n^2 + 1) - \ln(n^2 + 1) \). We can apply the logarithmic identity \( \ln(A) - \ln(B) = \ln\left(\frac{A}{B}\right) \) to simplify it. Therefore, \( a_n = \ln\left(\frac{2n^2 + 1}{n^2 + 1}\right) \).

Step 2: Inspect the limit expression

We analyze the expression inside the logarithm \( \frac{2n^2 + 1}{n^2 + 1} \). As \( n \rightarrow \infty \), the dominant terms are \( 2n^2 \) in the numerator and \( n^2 \) in the denominator. Thus, the expression simplifies to \( \frac{2n^2}{n^2} = 2 \).

Step 3: Evaluate the limit

Since the dominant terms give \( \frac{2n^2}{n^2} = 2 \), the expression inside the logarithm approaches 2 as \( n \rightarrow \infty \). Therefore, \( \ln\left(\frac{2n^2 + 1}{n^2 + 1}\right) \to \ln(2) \).

Step 4: Conclusion

The limit of the sequence \( a_n \) as \( n \rightarrow \infty \) is \( \ln(2) \). Since the limit exists, the sequence converges.

Key Concepts

Logarithmic Identities

Logarithmic identities are powerful tools in simplifying complex expressions involving logarithms. One of the most useful identities for sequences and series is the subtraction property:

• \( \ln(A) - \ln(B) = \ln\left(\frac{A}{B}\right) \)

This identity allows us to combine two logarithms into a single logarithm, which can often make calculations much simpler. For example, in the sequence \( a_n = \ln(2n^2 + 1) - \ln(n^2 + 1) \), applying this identity reduces it to \( a_n = \ln\left(\frac{2n^2 + 1}{n^2 + 1}\right) \).
This simplification is crucial as it sets up the expression for further evaluation, particularly when looking to find limits or determinations of convergence.

Limit Evaluation

Limit evaluation is the process of determining what, if anything, a sequence or function approaches as the variable grows indefinitely. In this exercise, we focus on the behaviour of the expression inside the logarithm:

• \( \frac{2n^2 + 1}{n^2 + 1} \)

To find this limit, assess the contribution of each term as \( n \) approaches infinity. Dominant terms usually dictate the expression's behaviour at infinity. Here, comparing the coefficients of the highest power terms is essential.
The key is simplifying the expression to see how it behaves in the limit. If the limit is a specific value, the sequence converges to that limit. In our case, the expression simplifies to \( 2 \), hinting that the sequence approaches some log value of \( 2 \).

Dominant Term Analysis

Dominant term analysis is a critical technique for simplifying expressions, especially when evaluating limits. When an expression contains multiple terms, some terms might overwhelm others as \( n \) grows larger.
In the expression \( \frac{2n^2 + 1}{n^2 + 1} \), notice that for large \( n \), the terms \( 2n^2 \) and \( n^2 \) make the most significant contributions since they increase at a quadratic rate.

• Ignore the \(+1\) terms because they become insignificant compared to \( n^2 \) as \( n \to \infty \).

Thus, the expression simplifies to \( \frac{2n^2}{n^2} = 2 \), which is straightforward once you focus on these dominant terms. This simple but effective strategy makes evaluating limits far more manageable.

Calculus Problem Solving

Successfully solving problems in calculus often involves a multi-step approach, using a combination of techniques. This particular exercise showcases such a method:

• Simplify the given expression using logarithmic identities.
• Evaluate the limit by analyzing dominant terms.
• Draw conclusions based on this simplification and computation.

Understanding how to apply these steps in sequence can unravel many calculus problems. Here, the convergence of \( a_n = \ln(2n^2 + 1) - \ln(n^2 + 1) \) was determined through precise application of identities, limit evaluation, and term analysis.
Mastering these strategic sequences can dramatically simplify problem solving in calculus, especially when tackling series and convergence issues.