Problem 69 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\left(\frac{3 n+1}{3 n-1}\right)^{n} $$

Step-by-Step Solution

Answer

The sequence converges to \( e^{2/3} \).

Step 1: Analyze the Limit of Inside Function

First, focus on the inside function \( b_n = \frac{3n+1}{3n-1} \). As \( n \to \infty \), the leading terms \( 3n \) dominate in both the numerator and the denominator. Thus, \( \frac{3n+1}{3n-1} \approx \frac{3n}{3n} = 1 \). Therefore, \( \lim_{n \to \infty} \frac{3n+1}{3n-1} = 1 \).

Step 2: Expression of the Sequence

The sequence is given by \( a_n = \left( \frac{3n+1}{3n-1} \right)^n \). We have established that \( \lim_{n \to \infty} \frac{3n+1}{3n-1} = 1 \). Thus, we are analyzing \( \lim_{n \to \infty} \left( 1 + \frac{2}{3n-1} \right)^n \).

Step 3: Apply the Exponential Limit

The expression \( \left( 1 + \frac{2}{3n-1} \right)^n \) can be approximated using the limit definition of the exponential function. Recall that \( \left( 1 + \frac{x}{n} \right)^n \to e^x \) as \( n \to \infty \). Set \( x = 2/3 \) for our sequence. Therefore, \( \left( 1 + \frac{2}{3n-1} \right)^n \approx e^{2/3} \).

Step 4: Conclusion of Convergence and Limit

Since as \( n \to \infty \), \( \left( \frac{3n+1}{3n-1} \right)^n \to e^{2/3} \), the sequence \( a_n \) converges. The limit of the sequence is \( e^{2/3} \).

Key Concepts

Limit of a Sequence

A sequence is a list of numbers in a specific order, and the limit of a sequence captures the behavior of these numbers as we progress toward infinity. Understanding the limit of a sequence is crucial for analyzing its long-term behavior. As you explore sequences, you often start by substituting larger and larger values into the sequence formula. For example, with the sequence given by \( a_n = \left( \frac{3n+1}{3n-1} \right)^n \), the first step is finding the limit of the inside function: \( b_n = \frac{3n+1}{3n-1} \).

• As \( n \to \infty \), the terms \(3n\) in both the numerator and denominator dominate, simplifying the expression to \( \frac{3n+1}{3n-1} \approx \frac{3n}{3n} = 1 \).
• Therefore, \( \lim_{n \to \infty} \frac{3n+1}{3n-1} = 1 \).

Recognizing this limit is the first move to tackling the question of whether a sequence tends toward a particular value (converges) or does not settle into a single value (diverges).

Convergence vs Divergence

In calculus, sequences can either converge or diverge. Convergence means the sequence approaches a specific value as it extends toward infinity. Divergence means the sequence grows indefinitely or oscillates without settling into a single value. The distinction helps us when determining the behavior of the given sequence \( a_n = \left( \frac{3n+1}{3n-1} \right)^n \).To evaluate convergence or divergence, consider whether a pattern or trend emerges:

• Convergence: The sequence hones in on a fixed number, like \( e^{2/3} \) in this case.
• Divergence: The sequence may grow without bound or constantly switch between values with no identifiable pattern.

By analyzing the given sequence, particularly by understanding the limit of its inner expression, we find that it hones in on \( e^{2/3} \), indicating convergence. Thus, the sequence \( a_n = \left( \frac{3n+1}{3n-1} \right)^n \) converges to \( e^{2/3} \).

Exponential Function Limits

Exponential limits come into play notably when sequences or functions involve expressions raised to a power that grows indefinitely. In the case of our sequence \( a_n = \left( \frac{3n+1}{3n-1} \right)^n \), we rely on the exponential function limit concept to determine convergence.Consider this principle: - If a sequence can be rewritten in the form \( \left( 1 + \frac{x}{n} \right)^n \), then as \( n \to \infty \), this approaches \( e^x \). Applying this to our sequence:

• The sequence transforms to \( \left( 1 + \frac{2}{3n-1} \right)^n \).
• As \( n \to \infty \), \( x \) is \( \frac{2}{3} \), so \( \left( 1 + \frac{2}{3n-1} \right)^n \approx e^{2/3} \).

Because the sequence can be approximated by such an exponential limit, this supports both the sequence's convergence and the specific limit value, \( e^{2/3} \). The understanding of exponential limits is key in analyzing the asymptotic behavior of sequences like this one.