Chapter 9: Problem 19
Converge, and which diverge? Use any method, and give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{\sin ^{2} n}{2^{n}}$$
Short Answer
Expert verified
The series \( \sum_{n=1}^{\infty} \frac{\sin^{2} n}{2^{n}} \) converges.
Step by step solution
01
Understanding the Series
We are given the series \( \sum_{n=1}^{\infty} \frac{\sin^{2} n}{2^{n}} \). To determine if it converges or diverges, we first observe the nature of the terms. The series is in the form of a power series with \( a_n = \frac{\sin^{2} n}{2^n} \).
02
Analyzing the Terms
The term \( \sin^2 n \) oscillates between 0 and 1 for all integer values of \( n \). Therefore, \( 0 \leq \sin^2 n \leq 1 \) for all \( n \). This means \( 0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n} \) for each term of the series.
03
Using the Comparison Test
To use the comparison test, we compare our given series to the geometric series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \). We know that \( 0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n} \) and the series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) converges because it is a geometric series with a common ratio \( r = \frac{1}{2} \) which is less than 1.
04
Applying the Comparison Test
Since \( \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n} \) and the series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) converges, the comparison test tells us that \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \) also converges. This is because a smaller or equal series than a converging series must also converge.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Comparison Test
The comparison test is a wonderful tool in the world of mathematics for determining the convergence or divergence of a series. Essentially, what this test does is compare a series of interest to another series whose behavior is already known. Here's how it works: if you have a series \( \sum a_n \) and you want to know if it converges, you find another series \( \sum b_n \) where \( 0 \leq a_n \leq b_n \) for all terms. If \( \sum b_n \) is a converging series, the comparison test tells us that \( \sum a_n \) must also converge.
In the exercise, the series \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \) is initially intimidating. But when breaking it down and comparing its terms to the terms of the geometric series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \), it becomes simpler. Since \( 0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n} \), and because the series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) is known to converge (more on that next), the comparison test lets us safely conclude that \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \) also converges.
In the exercise, the series \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \) is initially intimidating. But when breaking it down and comparing its terms to the terms of the geometric series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \), it becomes simpler. Since \( 0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n} \), and because the series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) is known to converge (more on that next), the comparison test lets us safely conclude that \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \) also converges.
Geometric Series
Geometric series are a class of series with terms that multiply by a constant factor each time. The standard form of a geometric series is \( \sum_{n=0}^{\infty} ar^n \) where \( a \) is the first term, and \( r \) is the common ratio. If \( |r| \lt 1 \), the geometric series converges to the sum \( \frac{a}{1-r} \). However, if \( |r| \geq 1 \), the series diverges.
- Understanding that it's because of this property that geometric series become useful in many convergence tests, such as the comparison test.
- The series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) used in the step-by-step solution is geometric, with a common ratio \( r = \frac{1}{2} \).
Oscillating Functions
In this exercise, oscillating functions come into play with the sine function, specifically \( \sin^2(n) \). An oscillating function is one that moves back and forth, typically between two values. For trigonometric functions like sine, this oscillation is periodic. The sine squared function, \( \sin^2(n) \), had a crucial role in our series.
- It oscillates between 0 and 1. This implies that for each integer \( n \), \( \sin^2(n) \) has varying values, yet remains bounded.
- Our interest in oscillating functions in the context of convergence relates to how these functions behave when placed under a converging series, like \( \frac{1}{2^n} \).