Problem 31 - Calculus: Early Transcendentals

Question

Use any test to determine whether the series is absolutely convergent, conditionally convergent, or divergent. \( \displaystyle \sum_{n = 2}^{\infty} \frac {( - 1)^n}{\ln n} \)

Step-by-Step Solution

Answer

The series is conditionally convergent.

Step 1: Determine Absolute Convergence

To check if the series is absolutely convergent, consider the series of the absolute values:\[ \sum_{n=2}^{\infty} \left| \frac{(-1)^n}{\ln n} \right| = \sum_{n=2}^{\infty} \frac{1}{\ln n}. \]The series \( \sum \frac{1}{\ln n} \) isn't a standard p-series or geometric series, so we need to compare it with a known divergent series or find its behavior. The integral test or limit comparison test can help establish the divergence of this series.

Step 2: Use Divergence Test for Absolute Series

Since \( \ln n \to \infty \) as \( n \to \infty \), the terms \( \frac{1}{\ln n} \) do not tend to zero fast enough. We can't directly apply the divergence test here, so instead, apply the integral test.Consider the integral:\[ \int_{2}^{\infty} \frac{1}{\ln x} \, dx. \]This integral diverges because the integral of \( \frac{1}{\ln x} \) from 2 to infinity is infinite. Therefore, by the integral test, \( \sum \frac{1}{\ln n} \) diverges. Hence, the original series is not absolutely convergent.

Step 3: Check Conditional Convergence Using Alternating Series Test

For the alternating series test, consider the original series:\[ \sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n}. \]This is an alternating series with \( b_n = \frac{1}{\ln n}. \)- Check if \( b_n \) is decreasing: Since \( \ln n \) is increasing for \( n \geq 2 \), \( b_n = \frac{1}{\ln n} \) is decreasing.- Check if \( \lim_{n \to \infty} b_n = 0 \): Yes, \( \lim_{n \to \infty} \frac{1}{\ln n} = 0. \)Both conditions of the alternating series test are satisfied, so the series is conditionally convergent.

Key Concepts

Alternating Series Test

The Alternating Series Test is a handy tool when dealing with series where the terms change signs. Let's consider the series \( \sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n} \). An alternating series can be formally described by an expression like \( \sum (-1)^n a_n \), where \( a_n \) is always positive. For the Alternating Series Test to confirm convergence, we need two key conditions:

• The sequence \( a_n \) must be decreasing. Essentially, each term should be smaller than or equal to the one before it. In our series, the term \( a_n = \frac{1}{\ln n} \) decreases as \( \ln n \) grows with \( n \).
• The limit of \( a_n \) as \( n \) approaches infinity should be zero. For our series, \( \lim_{n \to \infty} \frac{1}{\ln n} = 0 \).

When both these conditions are met, the series converges. In our example, since \( \frac{1}{\ln n} \) is a decreasing sequence and its limit is zero, the Alternating Series Test proves that our series is conditionally convergent.

Integral Test

The Integral Test helps to determine the convergence or divergence of a series by relating it to an improper integral. Take the series \( \sum_{n=2}^{\infty} \frac{1}{\ln n} \), which is the absolute value version of our original series. By performing an integral test, we look at the function \( f(x) = \frac{1}{\ln x} \), which needs to be positive, continuous, and decreasing for \( x \geq 2 \).
Set up the improper integral \( \int_{2}^{\infty} \frac{1}{\ln x} \, dx \). If this integral converges, then the series converges. Conversely, if the integral diverges, so does the series.
Upon evaluating, it turns out that the integral \( \int_{2}^{\infty} \frac{1}{\ln x} \, dx \) diverges. This means our series of absolute values, \( \sum \frac{1}{\ln n} \), also diverges, indicating that the original series is not absolutely convergent.

Conditional Convergence

Conditional Convergence arises when a series converges, but not absolutely. This means that while the original series converges, the series of its absolute values does not. Let's look at our series, \( \sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n} \).
We've already determined through the Alternating Series Test that it is convergent. But when considering the absolute values series, \( \sum \frac{1}{\ln n} \), we found that it diverges. This combination — where the series itself converges, but its absolute counterpart does not — is the hallmark of conditional convergence.
Conditional convergence reveals that the series relies on the interplay of positive and negative terms to "cancel out" and remain finite as \( n \) increases, offering a deeper insight into the behavior of alternating series.

Divergence Test

The Divergence Test is one of the simplest methods for checking a series for divergence. It states that if the limit of the sequence terms \( a_n \) as \( n \) approaches infinity is not zero, the series definitely diverges. However, if that limit equals zero, the test provides no conclusive result because the series may still either converge or diverge.
In our problem, checking the series \( \sum_{n=2}^{\infty} \frac{1}{\ln n} \), the term \( \frac{1}{\ln n} \) does indeed approach zero as \( n \) heads to infinity.
Since the limit is zero, we can’t conclude anything about convergence from the Divergence Test. That’s why other methods like the Integral Test are employed. Thus, while the Divergence Test is often the first step, it's not always the final say in determining a series' behavior.