Chapter 11: Problem 3
Which of the series in Exercises 1–36 converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{\sin ^{2} n}{2^{n}} $$
Short Answer
Expert verified
The series converges by the Comparison Test.
Step by step solution
01
Identify the Series
The given series is \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \). This is an infinite series with terms \( a_n = \frac{\sin^2 n}{2^n} \). The goal is to determine whether this series converges or diverges.
02
Consider the Properties of Terms
Note that \( \sin^2 n \) is always between 0 and 1 for any integer \( n \). This implies that \( 0 \leq \sin^2 n \leq 1 \). Therefore, each term of the series satisfies \( 0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n} \).
03
Compare with a Known Convergent Series
The series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) is a geometric series with a common ratio \( r = \frac{1}{2} \), which is less than 1. Therefore, it converges. Since each term of the given series is less than or equal to the corresponding term of this geometric series, \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \leq \sum_{n=1}^{\infty} \frac{1}{2^n} \), the given series also converges.
04
Conclusion Using Comparison Test
By the Comparison Test, if \( 0 \leq a_n \leq b_n \) for all \( n \) and \( \sum_{n=1}^{\infty} b_n \) converges, then \( \sum_{n=1}^{\infty} a_n \) also converges. In this case, because \( \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n} \) and \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) converges, it follows that \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \) converges.
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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 fundamental tool in determining the convergence or divergence of infinite series. It involves comparing a given series to another series whose convergence behavior is already known. Let’s simplify!Consider two infinite series, \( \sum a_n \) and \( \sum b_n \). The Comparison Test tells us:
- If \( 0 \leq a_n \leq b_n \) for all large \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
- Conversely, if \( a_n \geq b_n \geq 0 \) and \( \sum b_n \) diverges, then \( \sum a_n \) diverges too.
Geometric Series
A geometric series is one of the simplest forms of series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The basic form is:\[ \sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \dots \]Where:
- \( a \) is the first term.
- \( r \) is the common ratio.
Infinite Series
An infinite series is simply the sum of an infinite sequence of numbers. Imagine adding terms from a sequence: \( a_1, a_2, a_3, \ldots \), all the way to infinity. The sum, \( \sum_{n=1}^{\infty} a_n \), is what we call an infinite series. Infinite series can behave differently:
- **Convergent Series**: These series approach a specific number. They ‘settle down’ to a sum, no matter how many terms you add.
- **Divergent Series**: These series don’t settle to a fixed number. Instead, they either grow without bound or oscillate indefinitely.