Question

Explain why, for each \(n\), at least one of \(\\{|\sin n|,|\sin (n+1)|, \ldots,|\sin n+6|\\}\) is larger than \(1 / 2 .\) Use this relation to test convergence of \(\sum_{n=1}^{\infty} \frac{|\sin n|}{\sqrt{n}}\).

Short answer

At least one \(|\sin n|\) in any set of 7 is greater than 0.5, and \(\sum \frac{|\sin n|}{\sqrt{n}}\) diverges.

Step-by-step solution

Understanding the Sine Values

The sine function, \(|\sin x|\), is periodic with a period of \(2\pi\) and fluctuates between 0 and 1. Within any interval of \([n, n+6]\), the sine function completes a fraction of its period.

Properties of \(\sin x\) Range

The function \(\sin x\) attains every value in the range \([-1, 1]\) as \(x\) varies over any interval of length \(2\pi\). An interval of length 7 (from \(n\) to \(n+6\)) ensures that \(|\sin x|\) reaches above \(\frac{1}{2}\), because otherwise there would be segments where \(\sin x\) is flat without rising above \(\frac{1}{2}\), which isn’t possible over 7 consecutive points.

Guaranteeing Large \(|\sin x|\) Values

Consider any 7 consecutive integer points starting at \(n\). Since \(|\sin x|\) cannot remain below \(\frac{1}{2}\) for an entire "half-period" (\(\pi\)), at least one value in the set \(\{|\sin n|, |\sin(n+1)|, \ldots, |\sin(n+6)|\}\) must exceed \(\frac{1}{2}\).

Analyzing the Series for Convergence

To determine if \(\sum_{n=1}^{\infty} \frac{|\sin n|}{\sqrt{n}}\) converges, notice that for every seven consecutive terms, at least one term is larger than \(\frac{1}{2}\sqrt{n}\).

Showing Divergence

Utilizing the comparison test, observing that the terms do not decay more rapidly than \(\frac{1}{2}\sqrt{n}\) implies the comparison with the harmonic series \(\sum \frac{1}{n^{1/2}}\), which is known to diverge. Thus, \(\sum \frac{|\sin n|}{\sqrt{n}}\) diverges.

Key concepts

Sine Function Properties

The sine function, represented as \(\sin x\), is a fundamental trigonometric function that oscillates between -1 and 1. When considering the absolute value, \(|\sin x|\), the range is constrained between 0 and 1.
To explore this function further, we note the following properties:

• Periodicity: The sine function repeats its pattern every \(2\pi\) radians. This characteristic of repeating every complete cycle is known as its period.
• Symmetry: Due to its symmetry about the origin and axis, the sine function is considered odd. This means that \(\sin(-x) = -\sin(x)\).
• Critical Points: At multiples of \(\pi\), the sine function reaches zero, and at odd multiples of \(\frac{\pi}{2}\), it hits its maximum (1) or minimum (-1).

Understanding these properties helps us realize that within any interval that spans a reasonable portion of \(2\pi\), the function must inevitably reach a wide range of values. This ensures intervals like \([n, n+6]\) will contain sine values exceeding \(\frac{1}{2}\). For the task at hand, this is crucial in determining any patterns or properties related to convergence.

Comparison Test for Convergence

The comparison test is a useful tool in determining the behavior of infinite series. It allows us to infer the convergence or divergence of a given series by comparing it with another series whose behavior is already established.
Here's how it typically works:

• Finding a Series for Comparison: You first identify or construct another series, \(\sum b_n\), that you can compare with the original series, \(\sum a_n\).
• Applying the Test: You then check the terms \(a_n\) and \(b_n\):
  • If \(0 \leq a_n \leq b_n\) for all \(n\) and \(\sum b_n\) converges, then \(\sum a_n\) also converges.
  • If \(0 \leq b_n \leq a_n\) for all \(n\) and \(\sum b_n\) diverges, then \(\sum a_n\) also diverges.

In our specific example, the series \(\sum \frac{1}{n^{1/2}}\) is well-known to diverge (like a harmonic series). By showing that our terms \(\frac{|\sin n|}{\sqrt{n}}\) do not decrease more quickly than \(\frac{1}{\sqrt{n}}\), we conclude by the comparison test that the original series \(\sum \frac{|\sin n|}{\sqrt{n}}\) also diverges.

Periodicity of Functions

Periodicity is a key feature of many mathematical functions, especially trigonometric functions like sine and cosine. A function is periodic if it repeats its values in regular intervals, known as periods.
For the sine function, the interval of repetition, or period, is \(2\pi\). This property is pivotal when analyzing and predicting the behavior of such functions over repeated intervals.

• Understanding Repetition: Within each period from \(0\) to \(2\pi\), sine completes a full oscillation: starting at 0, reaching a maximum of 1 at \(\frac{\pi}{2}\), descending back to 0 at \(\pi\), dropping to a minimum of -1 at \(\frac{3\pi}{2}\), and finally returning to 0 at \(2\pi\).
• Predicting Values: Because of this periodic nature, knowing the sine value at any angle \(x\) means we also know it for \(x + 2k\pi\), where \(k\) is any integer.

In practical terms, periodicity allows us to predict and ensure that within any stretch of a few leading numbers like from \(n\) to \(n+6\), the sine values will vary sufficiently due to this repetition. This variability is crucial for evaluating the convergence of series that include trigonometric functions.