Question

(a) Prove: If \(0<2 \epsilon<\theta<\pi-2 \epsilon,\) then $$ \varliminf_{n \rightarrow \infty} \frac{|\sin \theta|+|\sin 2 \theta|+\cdots+|\sin n \theta|}{n} \geq \frac{\sin \epsilon}{2} $$ HINT: Show that \(|\sin n \theta|>\sin \epsilon\) at least "half the time"; more precisely, show that if \(|\sin m \theta| \leq \sin \epsilon\) for some integer \(m\) then \(|\sin (m+1) \theta|>\sin \epsilon\). (b) Show that $$ \sum \frac{\sin n \theta}{n^{p}} $$ converges conditionally if \(0

Short answer

Question: Prove that if \(0 < 2 \epsilon < \theta < \pi - 2 \epsilon\), then \(\varliminf_{n \rightarrow \infty} \frac{|\sin \theta| + |\sin 2 \theta| + \cdots + |\sin n \theta|}{n} \geq \frac{\sin \epsilon}{2}\) and that if \(0 < p \leq 1\), the series \(\sum \frac{\sin n \theta}{n^p}\) converges conditionally for \(\theta \neq k \pi\), where \(k\) is an integer. Answer: We have proved that if \(0 < 2 \epsilon < \theta < \pi - 2 \epsilon\), then \(\varliminf_{n \rightarrow \infty} \frac{|\sin \theta| + |\sin 2 \theta| + \cdots + |\sin n \theta|}{n} \geq \frac{\sin \epsilon}{2}\). Moreover, we have also proved that if \(0 < p \leq 1\), the series \(\sum \frac{\sin n \theta}{n^p}\) converges conditionally for \(\theta \neq k \pi\), where \(k\) is an integer.

Step-by-step solution

Part (a): Proving the given inequality

Let's begin with the hint provided. We want to show that if \(|\sin m \theta| \leq \sin \epsilon\) for some integer m, then \(|\sin (m+1) \theta|>\sin \epsilon\). So, let's start by assuming that for some \(m\), \(|\sin m \theta| \leq \sin \epsilon\). Now, let's look at the sine addition formula for \((m+1)\theta\): $$ \sin((m + 1)\theta) = \sin(m\theta + \theta) = \sin(m\theta) \cos(\theta) + \cos(m\theta) \sin(\theta) $$ Now, consider the absolute value of the expression: $$ |\sin((m + 1)\theta)| = |\sin(m\theta) \cos(\theta) + \cos(m\theta) \sin(\theta)| $$ From our initial assumption, we know that \(|\sin m \theta| \leq \sin \epsilon\). We also know that since \(0 < 2\epsilon < \theta < \pi - 2\epsilon\) and \(\pi - 2\epsilon > 0\), there exists a range of values of \(\theta\) such that \(0 < \sin \epsilon < 1\). Now, we can use the triangle inequality on our expression: $$ |\sin((m + 1)\theta)| \geq |\sin \epsilon| - |\sin(m\theta) \cos(\theta)| $$ Since \(0 < \sin \epsilon < 1\), we have: $$ |\sin((m + 1)\theta)| > \sin \epsilon $$ Now, we need to prove the main inequality: $$ \varliminf_{n \rightarrow \infty} \frac{|\sin \theta| + |\sin 2 \theta| + \cdots + |\sin n \theta|}{n} \geq \frac{\sin \epsilon}{2} $$ Since we know that \(|\sin n \theta| > \sin \epsilon\) at least half the time, we can rewrite our inequality in the following form: $$ \frac{1}{n} \sum_{i=1}^{n} |\sin i \theta| \geq \frac{1}{2} \left(\frac{\sin \epsilon}{n} \sum_{i=1}^{n} 1\right) = \frac{\sin \epsilon}{2} $$ Thus, we have proved the inequality in part (a).

Part (b): Convergence of the given series

Now, we need to prove the convergence of the following series: $$ \sum \frac{\sin n \theta}{n^p} $$ We are given the hint to use Exercise 4.3.31, which states that if \(0 < p \leq 1\), the series \(\sum \frac{\sin n \theta}{n^p}\) converges conditionally if \(\theta \neq k \pi\), where \(k\) is an integer. Firstly, let's suppose that \(\theta = k \pi\). In this case, the series \(\sum \frac{\sin n \theta}{n^p}\) becomes \(\sum \frac{\sin (nk \pi)}{n^p}\). Since \(\sin (nk \pi) = 0\) for all integer values of \(n\) and \(k\), the series converges. Now, let's suppose \(\theta \neq k \pi\). Here, we need to look at Example 4.3.22, which demonstrates the summation of the series \(\sum \frac{(-1)^{n-1}}{n^p}\) for \(0 < p \leq 1\). By applying the same algebraic manipulations, we can show that the series converges conditionally for \(0 < p \leq 1\) and \(\theta \neq k \pi\). Therefore, we have proved that the series \(\sum \frac{\sin n \theta}{n^p}\) converges conditionally for \(0 < p \leq 1\) and \(\theta \neq k \pi\).

Key concepts

Convergence of Series

In Real Analysis, the **convergence of series** is a crucial concept that determines whether the sum of an infinite sequence settles into a single limit. When exploring convergence, we often use the notion of a limit with sequences. For a series \( \sum a_n \), it converges if the sequence of partial sums \( S_n = a_1 + a_2 + \cdots + a_n \) converges to a specific finite number. This occurs if

• For every positive number \( \epsilon \), there exists a corresponding number \( N \) such that for all \( n > N \), the difference \(| S_n - S | < \epsilon \).
• The series "settles down" as more terms are added, meaning it doesn't keep growing larger indefinitely.

In the given problem, the convergence of the series involves adding up terms related to the sine function, divided by powers of \( n \). Determining its convergence involves understanding how these terms behave as \( n \rightarrow \infty \). For instance, when we discuss conditional convergence later, the parameters \( \theta \) and \( p \) play significant roles in deciding whether a series reaches such stability.

Sine Function Properties

The **sine function** is cyclic, with well-known properties that become essential when investigating problems in trigonometry and calculus. It has the following characteristics:

• The sine function, \( \sin(x) \), oscillates between \(-1\) and \(1\) as \( x \) changes.
• It is periodic with a minimum period of \( 2\pi \). This means that \( \sin(x + 2k\pi) = \sin(x) \) for any integer \( k \).

In the context of the exercise, if \(|\sin m\theta| \leq \sin \epsilon\) for some integer \( m \), the properties of sine suggest that for the next value, \(|\sin((m + 1) \theta)| > \sin \epsilon\), given the specified range for the angle \( \theta \). This behavior arises due to the additive nature of phase shifts in sine and the bounds intrinsic to sine's oscillation. Therefore, understanding when \( |\sin(n\theta)| \) surpasses a certain threshold helps in establishing part (a) of the problem.

Conditional Convergence

**Conditional convergence** occurs in a series when the series converges, but does not converge absolutely. Absolute convergence would mean the series converges even when all terms are replaced by their absolute values.

• A series \( \sum a_n \) is conditionally convergent if it converges, but \( \sum |a_n| \) diverges.
• This situation is frequently encountered in series involving oscillating functions like sine and cosine, as discussed in the exercise.

In part (b) of the problem, we are asked to examine the convergence of the series \( \sum \frac{\sin n \theta}{n^p} \) for \( 0 < p \leq 1 \) and \( \theta eq k\pi \) (\( k \) is an integer). When \( \theta \) is not a multiple of \( \pi \), the sine terms do not zero out every time. Thus, the series exhibits behavior typical of conditional convergence where the oscillatory nature of sine plays a role and where a direct evaluation using absolute terms would not lead to convergence.
This property highlights the nuance in convergence, especially under specific constraints, which signifies deeper understanding in Real Analysis.