Problem 60 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\sqrt[n]{n^{2}} $$

Step-by-Step Solution

Answer

The sequence converges, and its limit is 1.

Step 1: Understand the Sequence

The given sequence is \( a_n = \sqrt[n]{n^2} \), which can also be written as \( a_n = (n^2)^{1/n} \). Our task is to determine whether this sequence converges or diverges as \( n \rightarrow \infty \).

Step 2: Simplify the Expression

Rewrite the sequence \( a_n = (n^2)^{1/n} \) as \( a_n = n^{2/n} \). The goal is to analyze the behavior of \( a_n \) as \( n \to \infty \).

Step 3: Analyze the Limit

Consider \( a_n = n^{2/n} = e^{(2/n) \ln n} \). We need to find the limit of \( e^{(2/n) \ln n} \) as \( n \to \infty \).

Step 4: Apply Limit to the Exponent

Examine the expression \( \frac{2 \ln n}{n} \). As \( n \rightarrow \infty \), both \( \ln n \) and \( n \) tend to infinity, but \( n \) grows much faster. So, \( \frac{\ln n}{n} \to 0 \) as \( n \rightarrow \infty \). Therefore, \( \frac{2 \ln n}{n} \to 0 \).

Step 5: Conclude the Limit of the Sequence

Since the exponent \( \frac{2 \ln n}{n} \to 0 \), the expression \( e^{(2/n) \ln n} \to e^0 = 1 \). Hence, \( \lim_{n \to \infty} a_n = 1 \). This sequence converges to 1.

Key Concepts

Limits of Sequences

In mathematics, a sequence is a list of numbers in a specific order. The limit of a sequence describes the value that the elements of the sequence approach as the index becomes very large. This is particularly handy for understanding the behavior of infinite sequences, where the elements are ordered endlessly. To determine the limit, we analyze the sequence as the index goes to infinity.

When examining the sequence \(a_n = \sqrt[n]{n^2}\), we are interested in what happens as \(n\) approaches infinity. By simplifying it to \(a_n = n^{2/n}\), the structure reveals that the expression \frac{2 \ln n}{n}\ approaches 0 as \(n\) increases. Because \(e^{0} = 1\), the sequence converges to this value.

Understanding limits involves:

• Identifying an expression to simplify the sequence in terms that can be evaluated as \(n\) goes to infinity.
• Breaking down the expression carefully to recognize the behavior of its components over many terms.
• Concluding that if the expression approaches a constant value, the sequence converges; otherwise, it diverges.

Exponential Functions

Exponential functions are a type of mathematical function where the variable appears in the exponent. They are represented in the form \(e^{x}\), where \(e\) is the base of the natural logarithm (approximately 2.71828). These functions are characterized by their rapid growth or decay, which is dependant upon the sign and magnitude of the exponent.

In the sequence \(a_n = e^{(2/n) \ln n}\), the exponential function plays a crucial role in determining the limit. As we simplified \(a_n \,\text{to}\, e^{(2/n) \ln n}\), the function \(e^{x}\) depends on the behavior of the exponent \frac{2 \ln n}{n}\.

The exponential function \(e^{x}\) helps to elegantly capture growth trends in sequences:

• When the exponent approaches 0, as seen here, the function approaches 1, indicating convergence.
• When the exponent tends toward a positive or negative infinity, you get very large or very small results, depicting divergence.

Infinite Sequences

Infinite sequences are sequences that continue indefinitely. They consist of an ordered list of numbers that go on without ever stopping, getting successively larger or smaller. It is the task of identifying the long-term behavior of these sequences that enriches our understanding of mathematical structures.

In infinite sequences, the primary aim is to find if such sequences converge or diverge. Convergence means the terms in the sequence move towards a specific value. Divergence implies that the sequence does not settle towards a specific value.

Key aspects of infinite sequences include:

• Exploiting mathematical expressions to understand the long-run behavior.
• Applying the concept of limits to judge whether an infinite sequence converges or diverges.
• In our example of \(a_n = \sqrt[n]{n^2}\), after simplification and analysis, we observed that it converges to a limit of 1, demonstrating the power of calculating limits in assessing infinite sequences.