Problem 56 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=2}^{\infty} \frac{(\ln n)^{2}}{n} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=2}^{\infty} \frac{(\ln n)^{2}}{n} \) diverges.

Step 1: Identify the Type of Series

The given series is \( \sum_{n=2}^{\infty} \frac{(\ln n)^{2}}{n} \). This is a series where each term involves a logarithmic function raised to a power, divided by \( n \). Such series often resemble the p-series or can be compared with them.

Step 2: Comparison with p-Series

Recall a p-series \( \sum \frac{1}{n^p} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \). The given series has terms of the form \( \frac{(\ln n)^2}{n} \). This is not a traditional p-series but has a similar layout, suggesting the Limit Comparison Test could be useful here.

Step 3: Apply the Limit Comparison Test

Choose a comparison series that is easier to analyze. Here, use the harmonic series \( \sum \frac{1}{n} \), which diverges. Compute the limit:\[\lim_{n \to \infty} \frac{(\ln n)^2 / n}{1/n} = \lim_{n \to \infty} (\ln n)^2.\]Since \( (\ln n)^2 \) grows without bound as \( n \) goes to infinity, the limit is infinity.

Step 4: Interpret the Results

Since the limit is infinity and the harmonic series \( \sum \frac{1}{n} \) diverges, the Limit Comparison Test tells us that \( \sum \frac{(\ln n)^2}{n} \) also diverges.

Key Concepts

p-series comparison

The p-series is a fundamental concept in understanding the convergence and divergence of series. It takes the form \( \sum \frac{1}{n^p} \), where \( p \) is a positive constant. The key to analyzing a p-series is to determine the value of \( p \):

• If \( p > 1 \), the p-series converges.
• If \( p \leq 1 \), the p-series diverges.

In the original problem, we have a series \( \sum_{n=2}^{\infty} \frac{(\ln n)^{2}}{n} \). While this isn't a direct p-series (because of the \((\ln n)^2\) term), the structure prompts us to compare it to a p-series to understand its behavior. We need techniques like comparison tests to draw parallels to p-series, aiding in easier interpretation of convergence properties.

harmonic series

The harmonic series is a specific type of p-series where \( p = 1 \) and it is represented as \( \sum \frac{1}{n} \). This series is well-known for its divergence, even though its terms decrease as \( n \) increases. To see why it diverges, consider how the sum of its terms grows progressively over time without bound.In the exercise, the series \( \sum_{n=2}^{\infty} \frac{(\ln n)^{2}}{n} \) is compared with the harmonic series through the limit comparison test. Since the harmonic series diverges, it becomes a helpful tool when checking against other series, especially to determine divergence. If we can demonstrate similar behavior between the given series and the harmonic series, it indicates that the original series might also diverge.

limit comparison test

The Limit Comparison Test is a handy method when direct comparison with a base p-series or another known series is not straightforward. Suppose we have two series \( \sum a_n \) and \( \sum b_n \) with positive terms. To determine convergence or divergence using this test:

• Calculate \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
• If the limit is a positive finite number, both series either converge or diverge together.

In this exercise, the series \( \sum \frac{(\ln n)^2}{n} \) was compared to the harmonic series \( \sum \frac{1}{n} \), a diverging series, by applying the limit comparison test with these steps:

• Compute \( \lim_{n \to \infty} \frac{(\ln n)^2 / n}{1/n} \), which simplifies to \( \lim_{n \to \infty} (\ln n)^2 \).
• The result is infinity, indicating that \( \sum \frac{(\ln n)^2}{n} \) diverges, much like the harmonic series.

The test's strength lies in verifying similar behavior between complex series and simpler known ones, making it an essential tool in series analysis.