Problem 22 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 2}^{\infty} \frac {\ln n}{n^2} \)

Step-by-Step Solution

Answer

The series is convergent.

Step 1: Analyze Convergence Test Options

The series given is \( \sum_{n = 2}^{\infty} \frac{\ln n}{n^2} \). To determine its convergence, we can use either the Comparison Test or a similar known convergence test, such as the Limit Comparison Test. Here, we compare it with a series we suspect to be similar in behavior, particularly \( \sum_{n = 2}^{\infty} \frac{1}{n^2} \).

Step 2: Apply the Limit Comparison Test

The Limit Comparison Test requires examining \( \lim_{n \to \infty} \frac{\frac{\ln n}{n^2}}{\frac{1}{n^2}} = \lim_{n \to \infty} \ln n \). As \( n \rightarrow \infty \), \( \ln n \) also approaches \( \infty \), which means the limit is effectively \( \infty \). However, incorrectly applying this indicates divergence—we will adjust our approach.

Step 3: Refine the Comparison

Recognize that the initial approach needed adjustment, as \( \ln n \) growing to infinity diverges on its own. Instead, rewrite the comparison more generally: check if \( \frac{\ln n}{n^2} \leq \frac{n}{n^2} = \frac{1}{n} \) for sufficiently large \( n \). But note, \( \ln n < n \), hence, our previous results yield further insight.

Step 4: Directly Apply Known Convergent Series

Consider directly comparing with an easier convergent series \( \frac{1}{n^{3/2}} \), which converges (p-series with \( p > 1 \)). \( \frac{\ln n}{n^2} < \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}} \) holds true eventually after some threshold. Since \( \sum \frac{1}{n^{3/2}} \) converges, so does \( \sum \frac{\ln n}{n^2} \) by the Comparison Test.

Key Concepts

Comparison Test

The Comparison Test is a method to determine the convergence or divergence of series by comparing it with another series whose behavior we're familiar with. The core idea is:

• If a series \( \{ a_n \} \) can be compared to a series \( \{ b_n \} \) where \( a_n \leq b_n \) and \( \{ b_n \} \) converges, then \( \{ a_n \} \) also converges.
• Conversely, if \( a_n \geq b_n \) and \( \{ b_n \} \) diverges, then \( \{ a_n \} \) also diverges.

In the exercise, we used a modified approach of the Comparison Test. Initially, the function \( \frac{\ln n}{n^2} \) was compared with \( \frac{n}{n^2} = \frac{1}{n} \), but it was found not suitable for proving convergence directly. Instead, choosing \( \frac{1}{n^{3/2}} \) helps since it satisfies the inequality and is known to be a convergent series.

Limit Comparison Test

The Limit Comparison Test is useful when the direct Comparison Test is difficult to apply. This test involves calculating the limit:

• If \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \) and \( 0 < c < \infty \), then both series \( \sum a_n \) and \( \sum b_n \) either both converge or diverge.

In this exercise, there was a misstep initially while using this test. The limit \( \lim_{n \to \infty} \ln n \) goes to infinity. Hence, this approach initially seemed negative towards convergence. However, by refining the choice of \( b_n \), i.e., using \( \frac{1}{n^{3/2}} \), we ensure a suitable comparison towards convergence.

P-series

P-series are special mathematical series in the form \( \sum \frac{1}{n^p} \). The behavior of these series is well understood:

• The series converges if \( p > 1 \).
• The series diverges if \( p \leq 1 \).

This rule was essential in the exercise solution because \( \frac{1}{n^{3/2}} \) is a p-series with \( p = \frac{3}{2} > 1 \). Thus, it converges, and effectively at our disposal for comparing other series such as \( \frac{\ln n}{n^2} \) through the Comparison Test.

Convergent Series

A convergent series is one where the sum of its infinite terms approaches a finite number. It's a fundamental concept in calculus:

• Convergence indicates that as more terms are added, the series approaches some limit.
• Discovering if a series converges often involves tools like the Comparison Test or knowing specific types like p-series, geometric series, etc.

In the exercise, the original series \( \sum \frac{\ln n}{n^2} \) was shown to converge in part due to the knowledge of p-series and by leveraging the Comparison Test with the known convergent series \( \frac{1}{n^{3/2}} \). This showcases the interconnected nature of these mathematical concepts and the importance of understanding each to assess convergence efficiently.