Problem 10 - Thomas Calculus

Question

Determine if the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test. $$ \sum_{n=2}^{\infty}(-1)^{n+1} \frac{1}{\ln n} $$

Step-by-Step Solution

Answer

The series converges by the Alternating Series Test.

Step 1: Identify the series

The series given is \( \sum_{n=2}^{\infty}(-1)^{n+1} \frac{1}{\ln n} \). This is an alternating series because of the \((-1)^{n+1}\) factor, which causes the signs of the terms to alternate.

Step 2: Confirm the Alternating Series Test conditions

The Alternating Series Test requires that the terms \(b_n = \frac{1}{\ln n}\) are positive, decreasing, and approach zero as \( n \to \infty \). Let's check these conditions.

Step 3: Check if terms are positive

For \( n \geq 2 \), \( \ln n > 0 \), hence \( \frac{1}{\ln n} > 0 \). Therefore, the terms are positive.

Step 4: Check if terms are decreasing

To confirm that \(b_n = \frac{1}{\ln n}\) is decreasing, compare consecutive terms. Note that as \( n \) increases, \( \ln n \) also increases, which makes \( \frac{1}{\ln n} \) decrease. It can be rigorously demonstrated using calculus that \(\frac{d}{dn}\left(\frac{1}{\ln n}\right)<0\) for \(n>2\).

Step 5: Check the limit of the terms

Calculate \( \lim_{n \to \infty} \frac{1}{\ln n} = 0 \). As \( n \to \infty \), \( \ln n \to \infty \), thus \( \frac{1}{\ln n} \to 0 \).

Step 6: Conclusion using the Alternating Series Test

Since the terms are positive, decreasing, and approach zero, the series satisfies all the conditions of the Alternating Series Test. Therefore, the series \(\sum_{n=2}^{\infty} (-1)^{n+1} \frac{1}{\ln n}\) is convergent.

Key Concepts

Convergence of Series

Understanding the convergence of a series is crucial in determining whether the sum of infinite terms approaches a finite value. A series is essentially a sum of terms in a sequence, either finite or infinite. Convergence happens when, as you continue adding an infinite sequence, you approach a specific number.

For example, let's take the series \( \sum_{n=2}^{\infty}(-1)^{n+1} \frac{1}{\ln n} \). This is an **alternating series**, meaning each term in the sequence alternately subtracts and adds, because of the factor \((-1)^{n+1}\). The series converges if these alternating terms get steadily closer to zero without stopping.

The Alternating Series Test is a handy tool to determine this. A series \( \sum (-1)^n b_n \) converges if its positive terms \( b_n \) decrease and head towards zero. If these conditions are met, the alternating additions and subtractions pull the sum to a steady state, rather than growing endlessly or diverging chaotically.

Positive Decreasing Sequence

A system must meet specific criteria to pass the Alternating Series Test. One such essential requirement is that the sequence of terms should be a positive decreasing sequence.

In our series, terms are formed as \( \frac{1}{\ln n} \). These terms are positive because the natural logarithm of any number greater than 1 is positive, hence \( \ln n > 0 \) for \( n \geq 2 \).

Furthermore, to ensure it's a decreasing sequence, we verify that each term is smaller than the one preceding it. As \( n \) increases, \( \ln n \) also increases. This causes \( \frac{1}{\ln n} \) to become smaller as \( n \) gets larger. You can visualize this sequence getting smaller either by calculating specific terms or, more formally, by observing that the derivative \( \frac{d}{dn}\left(\frac{1}{\ln n}\right) < 0 \) for \( n > 2 \). This confirms the essential "decreasing term" condition for convergence in alternating series.

Calculus

Calculus plays a pivotal role in understanding and proving behaviors in series, especially when exploring the convergence and properties of sequences.

For our discussed series, calculus helps demonstrate the decreasing nature of the sequence by analyzing the function \( f(n) = \frac{1}{\ln n} \). By taking the derivative \( f'(n) \) and observing its sign, we confirm that the function is decreasing for \( n \geq 2 \).

Moreover, calculus aids in evaluating the limit of terms in a series; as \( n \to \infty \), \( \lim_{n \to \infty} \frac{1}{\ln n} = 0 \). This ability to determine behavior at infinity helps in meeting the condition of terms approaching zero in the Alternating Series Test.

• Understanding derivatives offers insights into how functions change.
• Limit evaluation is a core calculus tool for determining asymptotic behavior.

With these calculus techniques, we not only establish the conditions for convergence but also provide a more rigorous foundation for handling series.