Problem 37 - Thomas Calculus

Question

In Exercises \(31-38,\) use the \(n\) th-Term Test for divergence to show that the series is divergent, or state that the test is inconclusive. $$\sum_{n=1}^{\infty} \ln \frac{1}{n}$$

Step-by-Step Solution

Answer

The series diverges by the n-th term test.

Step 1: Identify the Sequence

For the series \( \sum_{n=1}^{\infty} \ln \frac{1}{n} \), we need to identify the sequence \( a_n \). In this case, the sequence is \( a_n = \ln \left( \frac{1}{n} \right) \), which simplifies to \( a_n = -\ln(n) \).

Step 2: Apply the n-th Term Test for Divergence

The \(n\)-th term test for divergence states that if \( \lim_{n \to \infty} a_n eq 0 \), then the series \( \sum_{n=1}^{\infty} a_n \) diverges. Calculate \( \lim_{n \to \infty} -\ln(n) \).

Step 3: Calculate the Limit

Determine \( \lim_{n \to \infty} -\ln(n) \). As \( n \to \infty \), \( \ln(n) \to \infty \), so \( -\ln(n) \to -\infty \). This limit is clearly not equal to zero.

Step 4: Conclude Divergence

Since \( \lim_{n \to \infty} a_n = -\infty eq 0 \), by the n-th term test, the series \( \sum_{n=1}^{\infty} -\ln(n) \) is divergent.

Key Concepts

Series Divergence

When dealing with infinite series, it's essential to determine whether a series converges to a finite value or diverges, meaning it doesn't settle to any particular value. The n-th Term Test, or the Test for Divergence, is a basic yet powerful tool used to identify diverging series.

The test is straightforward: if the limit of the sequence's terms isn't zero, the series definitely diverges. However, just because the term approaches zero doesn't mean the series converges. For example, the harmonic series is a well-known divergent series where its terms approach zero but the series itself does not converge.

In our given series,

• The limit of \( a_n = -\ln(n) \) as \( n \to \infty \) isn't zero. Instead, it approaches negative infinity, confirming divergence.
  

Knowing how to apply such tests to series helps you gauge their behavior quickly, an important skill in both calculus and real-world applications.

Sequence Identification

In any problem involving series, correctly identifying the underlying sequence is crucial. The sequence, denoted by \( a_n \), describes the individual terms you add up to form the series.

For our specific example, the series is \( \sum_{n=1}^{\infty} \ln \frac{1}{n} \). Here, the sequence \( a_n \) is \( \ln \left( \frac{1}{n} \right) \). With basic logarithm properties, you can rewrite this as \( a_n = -\ln(n) \).

This transformation simplifies further calculations and reveals patterns that might not be initially clear. Recognizing that \( \ln \left( \frac{1}{n} \right) \) converts to a negative logarithm is a vital step for further analysis.

• Proper sequence identification ensures correct application of mathematical tests.
• It unveils how each term contributes to the series' behavior.

Thus, mastering this identification paves the way for a better understanding of series behavior and further calculus operations.

Limit Calculation

Calculating the limit of a sequence is a decisive step in understanding its asymptotic behavior, especially when applying the n-th Term Test. In our example, we need to determine \( \lim_{n \to \infty} -\ln(n) \).

As \( n \) grows larger, \( \ln(n) \) increases without bound, approaching infinity. Hence, \( -\ln(n) \) approaches negative infinity.

• Key takeaway: \( a_n \) doesn't approach zero; it diverges to \( -\infty \).
• The non-zero limit confirms that the series diverges.

Grasping limits is fundamental not only for series but also for functions and their behavior at boundaries, aiding in robust calculus problem-solving skills.

Logarithmic Sequences

Logarithmic sequences often appear in mathematical series, posing intriguing behavior due to the logarithmic function's growth characteristics. Understanding these sequences requires comprehension of log properties.

The sequence in our exercise, \( \ln \left( \frac{1}{n} \right) \), simplifies to \( -\ln(n) \). This negative sign flips the direction of \( \ln(n) \), making it crucial to factor it into analysis.

• Logarithms grow slowly compared to polynomial functions, yet they can trend towards infinity.
• Negative logarithms, as seen here, trend towards negative infinity as \( n \) increases.
• Recognizing their behavior helps anticipate series and sequence patterns.

Logarithmic sequences demand a nuanced understanding due to their unique rate of increase. Mastering them leads to insightful mathematical exploration and problem-solving.