Problem 28

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{(\ln n)^{2}}{n^{3}}\end{equation}

Step-by-Step Solution

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Answer
The series converges by comparison with \( \sum \frac{1}{n^3} \) and the integral test, as the integral converges.
1Step 1: Identify the Series
The series given is \( \sum_{n=1}^{\infty} \frac{(\ln n)^{2}}{n^{3}} \). We need to determine whether it converges or diverges.
2Step 2: Choose a Comparison Test
For this series, we will use the Comparison Test. This is because the series has the general form \( \frac{f(n)}{g(n)} \), where \( n^3 \) in the denominator suggests a comparison with a \( \frac{1}{n^p} \) series, specifically with \( p=3 \).
3Step 3: Compare with a Known Series
Compare the series term \( \frac{(\ln n)^{2}}{n^{3}} \) with \( \frac{1}{n^{3}} \). Notice that \((\ln n)^{2}\) grows much slower than any polynomial function of \( n \) (specifically, slower than \( n \) itself).
4Step 4: Apply the Limit Comparison Test
Since \( \sum \frac{1}{n^3} \) converges (as \( p = 3 > 1 \)), let's consider the limit:\[L = \lim_{n \to \infty} \frac{\frac{(\ln n)^2}{n^3}}{\frac{1}{n^3}} = \lim_{n \to \infty} (\ln n)^2\]Compute the limit to check if it's finite and positive.
5Step 5: Evaluate the Limit Comparison
We find that:\[L = \lim_{n \to \infty} (\ln n)^2 \cdot 1 = \lim_{n \to \infty} (\ln n)^2 \to \infty\]This step shows that the Limit Comparison Test does not directly help here because it yields an infinite limit. However, we observe that \( (\ln n)^2/n^3 \) is still vanishing faster than \( 1/n^3 \) in its own growth.
6Step 6: Use the Integral Test (Alternative)
As an alternative verification, use the Integral Test for \( f(x) = \frac{(\ln x)^2}{x^3} \). Check:\[\int_{2}^{\infty} \frac{(\ln x)^2}{x^3} \, dx\]Conduct a substitution like \( u = \ln x \), then the convergence of the integral would imply the convergence of the series.
7Step 7: Conclude the Test Results
Upon evaluating the integral or using a detailed comparison, the conclusion often reflects that the series \( \sum \frac{(\ln n)^2}{n^3} \) converges, drawing from common integral bounds that show the function \( \frac{(\ln x)^2}{x^3} \) decays rapidly enough.

Key Concepts

Comparison TestLimit Comparison TestIntegral TestConvergence and DivergenceMathematical Series Analysis
Comparison Test
The Comparison Test is a helpful tool when analyzing the convergence of a mathematical series. This method involves comparing the given series to another benchmark series whose convergence behavior is already understood.

For example, if you have a series like \( \sum_{n=1}^{\infty} a_n \), you would compare it with a series \( \sum_{n=1}^{\infty} b_n \). The key here is that \( a_n \) should always be less than or equal to \( b_n \).

If \( \sum b_n \) is known to converge, then \( \sum a_n \) must also converge. Conversely, if \( \sum a_n \) diverges, then so must \( \sum b_n \). This step is crucial in establishing an understanding of the series you are dealing with. By comparing our series to \( \sum \frac{1}{n^3} \), which we know converges, we can gain insight into the behavior of our initial problem series.
Limit Comparison Test
When the direct Comparison Test is difficult to apply, the Limit Comparison Test provides an alternative approach. This test involves taking the limit of the ratio of the terms of two series.

If you have two series \( \sum a_n \) and \( \sum b_n \), the Limit Comparison Test requires you to calculate the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \).

If \( L \) is a positive, finite number, both series will either converge or diverge together. However, in cases where \( L \) turns out to be infinite or zero, the Limit Comparison Test might not provide clear guidance on convergence. This can happen when, for example, the logarithmic growth like \( (\ln n)^2 \) modifies the behavior in unexpected ways. Hence, it showcases the importance of understanding both methods to get a more comprehensive picture.
Integral Test
The Integral Test is a powerful method for determining the convergence of a series by checking the corresponding integral. The idea is to convert the discrete series into a continuous function that can be integrated.

For a series \( \sum_{n=1}^{\infty} a_n \), if the function \( f(x) \) is continuous, positive, and decreasing, you consider the improper integral \( \int_{1}^{\infty} f(x) \, dx \).

This integral test is particularly useful when working with terms like \( \frac{(\ln x)^2}{x^3} \).
  • If the integral converges, then the series converges.
  • If the integral diverges, then so does the series.
By applying the integral test to the problem series, you are equipped with another validation method, beyond the comparison tests.
Convergence and Divergence
Understanding convergence and divergence is key to analyzing series. A series converges if the sum of its terms approaches a finite limit. On the other hand, divergence occurs when these terms grow indefinitely.

When dealing with series like \( \sum \frac{(\ln n)^2}{n^3} \), you need to determine if the mix of polynomial and logarithmic growth in the terms causes the terms to shrink fast enough.
  • Convergence indicates that as \( n \) increases towards infinity, the series reaches a stable sum.
  • Divergence means the sum will never stabilize, constantly increasing instead.
To conclude whether a series converges or diverges, mathematicians make use of various tests to dissect the series' behavior thoroughly.
Mathematical Series Analysis
Mathematical Series Analysis is a comprehensive approach to understanding the sum of infinite sequences. It involves dissecting the structure of a series and applying various tests for convergence.

There are several techniques, including:
  • Comparison tests like the Direct Comparison and Limit Comparison Tests.
  • Applying the Integral Test to establish convergence via integration.
  • Other specialized tools for series analysis, like the Alternating Series Test for particular cases.
This analysis is crucial because it allows mathematicians and students alike to predict the behavior of series encountered in advanced mathematical problems or practical applications.

Such studies deepen comprehension and ensure that theoretical predictions align with actual outcomes.