Problem 63 - Essential Calculus

Question

Determine the convergence or divergence of the series. $$ \sum_{n=2}^{\infty} \frac{n}{\ln n} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the type of the series

This is an infinite series where each term is a function \(f(n) = \frac{n}{\ln n}\) of the term number \(n\). Such series can often be evaluated using the integral test or the comparison test.

Step 2: Apply the Integral Test

The integral test states that if \(f(n)\) is a positive, continuous, and decreasing function for all \(n\) beyond a certain point, and \(F(x)\) is an integral of \(f(x)\) from \(2\) to infinity, then \( \sum_{n=2}^{\infty} f(n)\) and \(F(x)\) either both converge or both diverge. If the integral \(F(x)\) diverges, then the sum also diverges. The integral of our function from \(2\) to infinity is given by \(F(x) = \int_{2}^{\infty} \frac{x}{\ln x} dx\). Solving this integral is non-trivial and does not result in a standard mathematical function, but it is known that this integral diverges.

Step 3: Conclusion

Since the integral \(F(x)\) diverges, it follows from the integral test that the series \( \sum_{n=2}^{\infty} \frac{n}{\ln n}\) also diverges.

Key Concepts

Infinite Series

When we talk about an infinite series, we're referring to the sum of the terms of an infinite sequence. The series Infinite series are fundamental concepts in the field of calculus and are particularly pertinent when dealing with functions and their behavior at infinity.

An infinite series is denoted as \(\sum_{n=a}^{\infty} f(n)\), where \('a'\) is the starting index and \('f(n)'\) is a function that provides the terms of the series. To determine whether the sum of an infinite series is finite (converges) or not (diverges), specific tests of convergence can be employed. The series in our problem, \(\sum_{n=2}^{\infty} \frac{n}{\ln n}\), is an example of such a series where the terms are generated by the function \(f(n) = \frac{n}{\ln n}\).

In the educational context, understanding whether a series converges or diverges is critical for further mathematical exploration and applications. For instance, convergence is essential to defining functions, evaluating integrals, and solving differential equations in advanced mathematics and engineering courses.

To assist with this, mathematicians have developed various tests for convergence, such as the integral test and the comparison test, which are used to analyze the behavior of a series. Understanding these tests requires recognizing the properties of the series in question—such as whether its terms are positive, decreasing, and continuous beyond a certain point.

Integral Test

The integral testis a method to determine the convergence or divergence of an infinite series. The test compares the series to a proper integral, which is often easier to evaluate. For the integral test to be used, the series must meet specific criteria: the function \(f(n)\) that generates the terms of the series must be positive, continuous, and decreasing for all sufficiently large \(n\).

Here's how the test works: consider the series \(\sum_{n=a}^{\infty} f(n)\) and the corresponding integral \(\int_{a}^{\infty} f(x) dx\). If the integral converges, meaning it has a finite value, then the series converges. Conversely, if the integral diverges, so does the series. This relationship stems from the idea that the area under the curve of \(f(x)\) from \(a\) to infinity gives us insight into the sum of the series.

In our exercise, we considered the series \(\sum_{n=2}^{\infty} \frac{n}{\ln n}\). The series meets the criteria for the integral test since \(\frac{n}{\ln n}\) is positive, continuous, and decreasing for all \(n > e\). When we evaluate the integral, we found that it diverges, which, by the integral test, leads to the conclusion that the series also diverges.

When creating educational content, it's vital to emphasize that while the integral test can be powerful, it might not always be straightforward to evaluate the corresponding integral. Sometimes it requires advanced techniques of integration or the acknowledgment that the integral's behavior is already known from mathematical theory.

Comparison Test

Another important tool in our convergence-testing toolkit is the comparison test. This test is particularly useful when the integral test is difficult to apply due to complex integration. The comparison test relies on comparing the series in question to a second series whose convergence is known.

The basic premise of the comparison test is simple: if we have two series, \(\sum a_n\) and \(\sum b_n\), where each term of \(a_n\) is less than or equal to \(b_n\), and \(\sum b_n\) converges, then \(\sum a_n\) also converges. Conversely, if every term of \(a_n\) is greater than or equal to \(b_n\), and \(\sum b_n\) diverges, then \(\sum a_n\) also diverges.

In our context, if the series of \(\frac{n}{\ln n}\) can be compared to a simpler series whose convergence properties are known, the comparison test could offer a faster route to determining convergence or divergence. However, the step-by-step solution provided does not apply the comparison test, possibly because finding an appropriate series for comparison is not straightforward.

When teaching students, we would also advise them to become familiar with classic comparison series, such as the p-series \(\sum \frac{1}{n^p}\), where the series converges if \(p > 1\) and diverges if \(p \leq 1\). This familiarity can help them rapidly assess a variety of series they might encounter in homework or exams. It's these fundamentals that can build a strong base for students as they progress through mathematical concepts.