Question

Show that the series converges. (a) \(\sum \frac{n \sin n \theta}{n^{2}+(-1)^{n}} \quad(-\infty<\theta<\infty)\) (b) \(\sum \frac{\cos n \theta}{n} \quad(\theta \neq 2 k \pi, k=\) integer \()\)

Short answer

Question: Determine if the following series converge or diverge: (a) \(\sum \frac{n \sin n \theta}{n^{2}+(-1)^{n}}\) (b) \(\sum \frac{\cos n \theta}{n}\) where \(\theta \neq 2k\pi\) Answer: (a) The series \(\sum \frac{n \sin n \theta}{n^{2}+(-1)^{n}}\) converges for all values of \(\theta\). (b) The series \(\sum \frac{\cos n \theta}{n}\) converges for all values of \(\theta\) not equal to \(2k\pi\).

Step-by-step solution

Part (a) - Apply the comparison test for convergence

Since sin function oscillates between -1 and 1, we can write: \(-1 \leq \sin n \theta \leq 1\). To show convergence, it is enough to show that the absolute term converges using comparison test. \(n|\sin n\theta| \leq n\) for all possible values of \(\theta\) and \(n\). The denominator \(n^2 + (-1)^n\) is always positive for any integer value of n. Now, we can bound the absolute term: \(0\leq \frac{n|\sin n\theta|}{n^2+(-1)^n}=\frac{n|\sin n\theta|}{n^2+(-1)^n} \leq \frac{n}{n^2+1}\) To determine the convergence, let's examine the series \(\sum \frac{n}{n^2+1}\) using the comparison test. Since \(\frac{n}{n^2+1} \leq \frac{n}{n^2}=\frac{1}{n}\), we know that this series converges by integral test. Thus, by the comparison test, the original series \(\sum \frac{n \sin n \theta}{n^{2}+(-1)^{n}}\) converges for all values of \(\theta\).

Part (b) - Apply the alternating series test for convergence

Let's write the given series as an alternating series by multiplying both the numerator and denominator by \((-1)^n\). The new series can be written as: \(\sum \frac{(-1)^n \cos n \theta}{n}\) Now we can apply the alternating series test for convergence. 1. Verify the sequence \(\{a_n\} = \{\frac{|\cos n \theta|}{n}\}\) is decreasing 2. Check if \(\lim_{n\to\infty} a_n = 0\) The term \(a_n = \frac{|\cos n \theta|}{n}\) is decreasing since the cosine function is bounded between -1 and 1. As \(n\) increases, the term \(\frac{|\cos n \theta|}{n}\) will reduce in magnitude. Now, let's check for the limit as \(n\) approaches infinity: \(\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{|\cos n \theta|}{n} = 0\) Since both conditions are satisfied i.e., sequence is decreasing and its limit goes to 0, by the alternating series test, the original series \(\sum \frac{\cos n \theta}{n}\) converges for all values of \(\theta\) not equal to \(2k\pi\).

Key concepts

Comparison Test

The comparison test is a powerful tool used to determine the convergence or divergence of series by comparing them to a series whose convergence status is known. In this exercise, Part (a) requires us to apply the comparison test to prove that the series \[ \sum \frac{n \sin n \theta}{n^{2}+(-1)^{n}} \]converges. The sine function, \( \sin n \theta \), oscillates between -1 and 1 for all values of \( n \) and \( \theta \), making it ideal for the comparison test.
Since the absolute value \( |\sin n \theta| \leq 1 \), we bound the original series to:

• \( 0 \leq \frac{n| \sin n \theta|}{n^2+(-1)^n} \leq \frac{n}{n^2+1} \)

We know that \( \sum \frac{1}{n} \) diverges, but the \( \sum \frac{n}{n^2} \) converges using the integral test. Because the series \( \sum \frac{n}{n^2+1} \) is smaller than a known convergent series, by comparison, the original series also converges for all \( \theta \).
This method gives insight into how functions can be bounded to meet the threshold of convergence. Thus, by applying the comparison test efficiently, we affirm the convergence of a given series.

Alternating Series Test

In Part (b) of the exercise, we explore the alternating series test, which is another method to determine convergence—specifically for series that alternate in sign. The original series given is:\[ \sum \frac{\cos n \theta}{n} \]For the test to be applicable, the series is expressed with alternating terms by multiplying the numerator by \((-1)^n\):

• \(\sum \frac{(-1)^n \cos n \theta}{n}\)

Understanding the alternating series test involves checking two main conditions:

• The absolute sequence \( \{a_n\} = \left\{\frac{|\cos n \theta|}{n}\right\}\) must be decreasing.
• \( \lim_{n\to\infty} a_n = 0 \)

Given the bounded nature of the cosine function such that \(|\cos n \theta| \leq 1\), the quantity \( \frac{|\cos n \theta|}{n} \) diminishes with increasing \( n \), satisfying both decreasing and limit conditions. Therefore, both criteria for the alternating series test are met, ensuring the series converges for all given \( \theta eq 2k\pi \).
This highlights the simplicity yet robust applicability of the alternating series test in proving the convergence of series with alternating behavior.

Sin and Cos Functions

Understanding the behavior of trigonometric functions like \( \sin \) and \( \cos \) is critical for applying convergence tests effectively. Both these functions are periodic and oscillate between -1 and 1. This bounded nature makes them significant in convergence proofs.

The sine function, \( \sin n \theta \), has no fixed period when \( \theta \) varies, as its phase changes with \( n \) and \( \theta \). However, its values remain well-contained within [-1, 1].

• This feature is handy when comparing or bounding series terms.

Similarly, \( \cos n \theta \) behaves periodically and is used in analyzing series that may alternate in sign, adding another layer of consideration in convergence analysis. For instance, in an alternating series where \( \cos \)is utilized, the function's limits have a pivotal role in satisfying convergence test criteria.

These characteristics make sine and cosine functions valuable tools in mathematical analysis, especially when coupled with convergence principles to solve complex series-related problems.