Problem 11 - Calculus Single Variable

Question

The given series may be shown to converge by using the Alternating Series Test. Show that the hypotheses of the Alternating Series Test are satisfied. $$ \sum_{n=2}^{\infty} \cos (\pi n) \sin (\pi / n) $$

Step-by-Step Solution

Answer

The series \(\sum_{n=2}^{\infty} \cos(\pi n) \sin(\pi/n)\) converges by the Alternating Series Test.

Step 1: Identify the Alternating Series Components

The given series is \(\sum_{n=2}^{\infty} \cos (\pi n) \sin (\pi / n)\). First, notice that \(\cos(\pi n)\) alternates between 1 and -1 based on whether \(n\) is even or odd, making the series alternating. Thus, we can write the series as \(\sum_{n=2}^{\infty} (-1)^n b_n\) where \(b_n = \sin(\pi / n)\).

Step 2: Check if Terms Decrease to Zero

To apply the Alternating Series Test, we need to verify that \(b_n = \sin(\pi / n)\) is a decreasing sequence and that \(\lim_{n \to \infty} b_n = 0\). As \(n\) increases, \(\pi/n\) decreases, and \(\sin(\pi/n)\) approaches zero. Thus, \(\lim_{n \to \infty} \sin(\pi/n) = \sin(0) = 0\).

Step 3: Verify Monotonic Decrease of Terms

To verify the monotonic decrease of \(b_n = \sin(\pi/n)\), observe that the derivative of \(\sin(\pi/x)\) with respect to \(x\) is negative for \(x > 0\), indicating that \(\sin(\pi/n)\) is decreasing as \(n\) increases.

Step 4: Conclusion on Alternating Series Test

Since both hypotheses of the Alternating Series Test are satisfied — the terms \(b_n = \sin(\pi/n)\) are decreasing and approach 0 — we can conclude that the series \(\sum_{n=2}^{\infty} \cos(\pi n) \sin(\pi/n)\) converges.

Key Concepts

Convergent Series

A convergent series is a sequence of numbers that sums to a specific finite value as the number of terms approaches infinity. Unlike divergent series, which grow indefinitely, convergent series settle on a particular number. It is important to know that not all infinite series converge. The behavior of the terms and their limits play crucial roles.
For instance, the alternating series in our exercise converges because its terms decrease in absolute magnitude and tend to zero. The Alternating Series Test helps in verifying this by considering two main conditions:

• The sequence of terms is strictly decreasing.
• The limit of the terms as they go to infinity is zero.

These conditions are satisfied in the given series, proving it to be convergent. The beauty of convergent series lies in their predictable nature and the ability to approximate real-world quantities.

Monotonic Decreasing Sequence

A monotonic decreasing sequence is a sequence where each term is less than or equal to the previous term. In simpler terms, as you proceed through the sequence, the values either remain constant or decrease. These types of sequences are pivotal in mathematical analysis due to their predictable and controlled behavior.
In the exercise provided, the sequence given by the terms \(b_n = \sin(\pi/n)\) is a perfect example of a monotonic decreasing sequence. As \(n\) increases, the fraction \(\pi/n\) becomes smaller, causing \(\sin(\pi/n)\) to approach zero from above. This follows from the basic property of the sine function, where smaller angles result in smaller sine values.
To conclusively determine that \(\sin(\pi/n)\) is decreasing, you notice that its derivative with respect to \(n\) is negative, confirming the decreasing trend. Such sequences are essential in confirming the convergence of alternating series, as shown in this exercise.

Trigonometric Series

A trigonometric series is an infinite series formed by trigonometric functions like sine and cosine. These series are fundamental in mathematics and applications like signal processing. The Alternating Series Test is often used to determine convergence in these cases, especially when the series alternates between positive and negative terms due to the cosine function.
In our exercise, the series \(\sum_{n=2}^{\infty} \cos(\pi n) \sin(\pi/n)\) is a trigonometric series that alternates. This is because \(\cos(\pi n)\) takes the values \(1\) and \(-1\) alternatingly as \(n\) changes from even to odd.
The sine function in the series assists in driving the terms toward zero. Together, these trigonometric aspects define the series's structure and decide its behavior. Understanding how each part - sine and cosine functions - influence the series is crucial for analyzing and predicting convergence.