Problem 63 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The sequence \(a_n = \frac{\ln n}{n^{1/n}}\) diverges as \(n\) approaches infinity.

Step 1: Analyze the Denominators

First, observe the denominator of the sequence: \(n^{1/n}\). As \(n\) approaches infinity, the expression \(n^{1/n}\) is known to converge to 1 since for large values of \(n\), \(n^{1/n} = e^{\ln(n)/n}\) and \(\ln(n)/n\) approaches zero.

Step 2: Analyze the Numerator Function

Next, examine the numerator: \(\ln n\). The natural logarithm function \(\ln n\) increases without bound as \(n\) approaches infinity, though it does so very slowly.

Step 3: Apply L'Hôpital's Rule

The sequence \(a_n = \frac{\ln n}{n^{1/n}}\) can be analyzed using L'Hôpital's Rule, since the form is \(\frac{\infty}{1}\) as \(n\to \infty\). However, instead of using L'Hôpital's, we can directly note that since \(n^{1/n}\to 1\), the sequence \(\frac{\ln n}{n^{1/n}}\) behaves approximately as \(\ln n\) for large \(n\), which diverges.

Step 4: Conclusion on Convergence/Divergence

In conclusion, given that the numerator \(\ln n\) diverges to infinity faster than the denominator approaches 1, the sequence \(a_n\) diverges to infinity.

Key Concepts

Limit of a Sequence

To understand whether a sequence converges or diverges, we need to find out the behavior of its terms as the sequence progresses. The sequence in question is \( a_n = \frac{\ln n}{n^{1/n}} \). When we talk about convergence, we mean that as \( n \) gets larger and larger, the sequence's terms approach a specific value, known as the limit.

For a sequence to converge, the limit should exist and be finite, meaning the terms get closer and "stick" to a number. If a sequence does not converge to a specific number, it diverges.

In our case, examining the terms of the sequence \( \frac{\ln n}{n^{1/n}} \), we need to consider both the numerator and denominator separately. This understanding forms the basis of determining the ultimate behavior of the sequence.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a way to express logarithmic relationships with the base \( e \), where \( e \approx 2.71828 \). It tells us how many times we need to multiply \( e \) to get \( x \).

• The function \( \ln x \) is defined for all positive values of \( x \) and increases as \( x \) becomes larger.
• However, it increases very slowly compared to other functions, like linear or exponential functions.

In the sequence \( a_n = \frac{\ln n}{n^{1/n}} \), the natural logarithm is in the numerator, and as \( n \) approaches infinity, \( \ln n \) also approaches infinity. This slow but unbounded growth is vital as it determines part of the sequence's limit behavior in large \( n \).

Knowing that \( \ln n \) grows indefinitely helps us anticipate the behavior of sequences where \( \ln n \) is a component.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to find limits of indeterminate forms, notably \( 0/0 \) or \( \infty/\infty \). If a function presents this structure, L'Hôpital's Rule allows us to take derivatives of the numerator and the denominator until we can evaluate the limit.

In this exercise, the form \( \frac{\ln n}{n^{1/n}} \) becomes increasingly similar to \( \frac{\infty}{1} \) as \( n \to \infty \). Despite this, analyzing without L'Hôpital's Rule shows \( n^{1/n} \to 1 \), simplifying the sequence to reflect mainly the behavior of \( \ln n \). In essence, L'Hôpital's Rule is avoided here because the denominator sufficiently approaches 1.

However, when faced with similar indeterminate expressions, L'Hôpital's Rule often steps in, offering a pathway to simplifying complex limit problems by transforming them into solvable derivatives.