Question

Use the Direct Comparison Test to determine the convergence of the given series; state what series is used for comparison. $$\sum_{n=2}^{\infty} \frac{1}{n^{2} \ln n}$$

Short answer

The series \( \sum_{n=2}^{\infty} \frac{1}{n^2 \ln n} \) converges by comparison to the convergent p-series \( \sum \frac{1}{n^2} \).

Step-by-step solution

Understand the Series

The series we are given to analyze is \( \sum_{n=2}^{\infty} \frac{1}{n^2 \ln n} \). We aim to determine whether this series converges or diverges using the Direct Comparison Test.

Choose a Comparison Series

To use the Direct Comparison Test, we need to compare our series to a known convergent or divergent series. The series \( \sum_{n=2}^{\infty} \frac{1}{n^2} \) is a good candidate because it is a p-series with \( p = 2 \), which converges. Since \( \ln n > 1 \) for all \( n \geq 2 \), we find that \( \frac{1}{n^2 \ln n} < \frac{1}{n^2} \).

Apply the Direct Comparison Test

According to the Direct Comparison Test, if \( 0 \leq a_n \leq b_n \) for all \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) converges. In this case, let \( a_n = \frac{1}{n^2 \ln n} \) and \( b_n = \frac{1}{n^2} \). Since \( 0 \leq \frac{1}{n^2 \ln n} \leq \frac{1}{n^2} \) for \( n \geq 2 \) and \( \sum \frac{1}{n^2} \) converges, \( \sum \frac{1}{n^2 \ln n} \) also converges.

Conclude the Analysis

Since we have established that \( \frac{1}{n^2 \ln n} < \frac{1}{n^2} \) and the series \( \sum \frac{1}{n^2} \) is convergent, by the Direct Comparison Test, the series \( \sum \frac{1}{n^2 \ln n} \) must also converge. Thus, the given series converges.

Key concepts

Convergence Tests

Convergence tests are essential tools in calculus to determine whether a series converges or diverges. When dealing with infinite series, convergence implies that as the series progresses indefinitely, it approaches a finite limit. Divergence, on the other hand, suggests the series does not settle at a limit. One of the widely used methods is the Direct Comparison Test. This test involves comparing two series term by term. If you can show that all terms of your series of interest are less than or equal to a corresponding known convergent series, then your series converges. Conversely, if all terms are greater than or equal to a known divergent series, then your series diverges. The key is selecting an appropriate series for comparison, often a simpler one whose behavior is already established.

P-Series

A p-series is a series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^{p}} \). Whether a p-series converges or diverges depends on the value of \( p \). Specifically:

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

In the provided exercise, the series \( \sum_{n=2}^{\infty} \frac{1}{n^2} \) is used as a comparison. Here, \( p = 2 \), indicating convergence. This makes p-series an excellent choice for comparisons in the Direct Comparison Test, as their convergence properties are well understood. Expanding the use of p-series helps in efficiently deciphering the behavior of more complex series.

Series Comparison

Series comparison is a strategic approach employed when dealing with series where determining convergence directly is challenging. By comparing the given series to another series with known behavior, one can infer the convergence or divergence of the first series. The process involves strategically choosing a comparison series; ideally, a simpler series, like a p-series, that already has known convergence properties.

• Identify if the terms are greater or smaller than the known series.
• Apply the Direct Comparison Test according to the relation between the terms.

The series from the exercise, \( \frac{1}{n^2 \ln n} \), was compared with the simpler \( \frac{1}{n^2} \). Since \( \ln n > 1 \) for \( n \geq 2 \), the terms \( \frac{1}{n^2 \ln n} < \frac{1}{n^2} \). This relationship aids in leveraging the known convergence property of the comparison series to conclude the behavior of the original series.

Logarithmic Functions

Logarithmic functions, often represented as \( \ln(n) \), play a critical role in series analysis, especially when they appear in the terms of the series. In the examined series, the denominator, \( n^2 \ln n \), features the logarithmic function, indicating its potential influence on convergence behavior.Understanding how logarithms behave is crucial.

• Logarithmic growth is slower than any power of \( n \) for large values of \( n \).
• This property means that in series like \( \frac{1}{n^2 \ln n} \), the logarithm may help the series converge faster when compared to \( \frac{1}{n^2} \).

In our exercise, for \( n \geq 2 \), \( \ln n \) is always greater than 1, reducing the impact of \( n^2 \ln n \) relative to \( n^2 \) in the denominator. Hence, the terms of the series \( \frac{1}{n^2 \ln n} \) remain smaller, facilitating its comparison and conclusion of convergence through the Direct Comparison Test.