Problem 61 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Sequence converges; limit is 1.

Step 1: Understand the Problem

We are given a sequence \(a_n = \left(\frac{3}{n}\right)^{1/n}\) and need to determine whether it converges or diverges. If it converges, we will also find its limit.

Step 2: Evaluate the Sequence as n Approaches Infinity

To decide the convergence of the sequence, investigate \( \lim_{{n \to \infty}} a_n = \lim_{{n \to \infty}} \left(\frac{3}{n}\right)^{1/n} \). This expression suggests considering how \( \frac{3}{n} \) behaves and how the exponent \( \frac{1}{n} \) affects this.

Step 3: Simplify and Analyze the Sequence

Recognize that as \( n \) goes to infinity, \( \frac{3}{n} \to 0\). The sequence's form \( x^{1/n} \) for \( x \approx 0 \) behaves like 1. More precisely, since \( a_n = e^{\frac{1}{n} \ln(\frac{3}{n})}\), analyze \( \ln(\frac{3}{n}) = \ln(3) - \ln(n) \) as \( n \to \infty \).

Step 4: Apply Logarithm and Exponential Properties

Observe that \( \frac{\ln(3) - \ln(n)}{n} = \frac{\ln(3)}{n} - \frac{\ln(n)}{n} \). As \( n \to \infty, \frac{\ln(3)}{n} \to 0 \) and \( \frac{\ln(n)}{n} \to 0 \) by the definition of limits resulting from L'Hopital's Rule on \( \frac{\ln(n)}{n} \).

Step 5: Conclusion on Convergence and Limit

Since both terms in \( \frac{\ln(3)}{n} - \frac{\ln(n)}{n} \) approach 0, the whole expression converges to 0. Therefore, \( e^{0} = 1 \). Hence, \( a_n = \left(\frac{3}{n}\right)^{1/n} \) converges to 1.

Key Concepts

Sequence Limits

The concept of sequence limits is a foundational topic in calculus. When we talk about limits in the context of sequences, we're examining whether a sequence of numbers approaches a particular value as we move towards infinity. If such a value exists, the sequence is said to converge. If not, it diverges.
The limit of a sequence \( \{a_n\} \) as \( n \) goes to infinity is written as \( \lim_{{n \to \infty}} a_n \). This means we're interested in the behavior of \( a_n \) for very large values of \( n \.\)

• Converging sequences approach a specific number (limit).
• Diverging sequences do not settle on a single value.

In the given exercise, you have \( a_n = \left(\frac{3}{n}\right)^{1/n} \). The task is to determine if this sequence converges and, if so, find its limit. By analyzing \( \left(\frac{3}{n}\right)^{1/n} \) as \( n \) approaches infinity, we deduce whether it's becoming stable at a single point.

Exponential Functions

Exponential functions are a type of mathematical function where the variable appears in the exponent. They are denoted as \( e^x \), where \( e \) is a constant approximately equal to 2.71828, known as Euler's number. These functions have interesting properties, particularly when dealing with behavior at infinity.
In the context of the exercise, the sequence \( a_n = e^{\frac{1}{n} \ln(\frac{3}{n})}\) showcases such properties. This transformation arises from rewriting powers in an exponential form using the natural logarithm:
\[\left(\frac{3}{n}\right)^{1/n} = e^{\frac{1}{n} \ln(\frac{3}{n})}\]
Exploring this form helps us understand the interaction of the exponential function with limits. Thanks to the properties of logs and exponentials:

• As \( n \to \infty, \ln(\frac{3}{n}) \) becomes \( \ln(3) - \ln(n) \).
• The term \( \frac{1}{n} \ln(\frac{3}{n}) \) approaches 0, leading the whole expression inside the exponent to approach 0.
• Thus, the exponential of 0 is 1, demonstrating that this transformation helps find limits of complex forms.

L'Hopital's Rule

L'Hopital's Rule is a valuable tool used in calculus to determine limits of indeterminate forms such as \( \frac{0}{0}\) or \( \frac{\infty}{\infty}\). This rule states that if the limits of both the numerator and denominator approach 0 or infinity, the limit of the fraction can be found by differentiating the numerator and the denominator separately and then evaluating the limit.
In the given exercise, determining the limit of \( \frac{\ln(n)}{n} \) as \( n \) approaches infinity is an example of applying L'Hopital's Rule:

• Consider \( \lim_{{n \to \infty}} \frac{\ln(n)}{n} \).
• Both the numerator \( \ln(n) \) and denominator \( n \) approach infinity, fitting the criteria for L'Hopital's Rule.
• Differentiate both to get: the derivative of \( \ln(n) \) is \( \frac{1}{n} \) and of \( n \) is 1.
• Then evaluate \( \lim_{{n \to \infty}} \frac{1}{n} = 0 \).

This simplifies the analysis needed to determine that \( \frac{\ln(3)}{n} - \frac{\ln(n)}{n} \) results in a limit of 0, allowing us to conclude that the entire expression converges to 1. L'Hopital's Rule is instrumental in breaking down the complexity of limits in calculus.