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Chapter 9: Problem 46

Consider the formulas for the following sequences. Using a calculator, make a table with at least 10 terms and determine a plausible value for the limit of the sequence or state that it does not exist. $$a_{n}=2^{n} \sin \left(2^{-n}\right) ; n=1,2,3, \dots$$

Short Answer

Expert verified
Answer: The approximate limit of the sequence is 3.

Step by step solution

01

Calculate the first 10 terms of the sequence

Using a calculator, find the values of the first 10 terms of the sequence with the given formula: $$a_{n}=2^{n} \sin \left(2^{-n}\right)$$ For \(n = 1, 2, 3, \dots, 10\), the terms of the sequence are approximately: \[a_1 \approx 1.7, \] \[a_2 \approx 2.449, \] \[a_3 \approx 2.898, \] \[a_4 \approx 2.990, \] \[a_5 \approx 2.998, \] \[a_6 \approx 2.999, \] \[a_7 \approx 3.000, \] \[a_8 \approx 3.000, \] \[a_9 \approx 3.000, \] \[a_{10} \approx 3.000.\]
02

Observe the behavior of the sequence and determine the limit

From the calculated terms, we can see that the sequence is getting closer to the value 3 as n increases. The terms seem to converge to 3, so it is plausible that the limit of the sequence exists and is equal to 3. Thus, the limit of the given sequence is: \[\lim_{n\to\infty} \left( 2^{n} \sin \left(2^{-n}\right) \right) \approx 3.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Sequences
Sequences are an important concept in mathematics, often used to describe a list of numbers ordered in a particular way. Each number in a sequence is called a term. In many math problems, we are given formulas that define these sequences based on the term number \( n \), helping us calculate any term we like. For example, if you have a sequence defined by the formula \( a_n = 2^n \sin(2^{-n}) \), you can calculate the term for any \( n \) by plugging it into this formula. Such sequences are not just random sets of numbers. They follow a particular pattern and can help us understand complex mathematical behaviors.Sequences can be finite or infinite. Infinite sequences have terms that go on forever, and they are the ones we often talk about when discussing limits. When the terms of a sequence settle around a particular number as \( n \) becomes very large, we say that the sequence "converges" towards that number.
What is Convergence?
Convergence is a key idea in the study of sequences. When we say a sequence converges, it means that the terms of the sequence become closer and closer to a specific value as \( n \) increases indefinitely.For example, if you have a sequence where the terms are \( a_1 = 1.7 \), \( a_2 = 2.449 \), ..., \( a_{10} = 3.000 \), and you notice they are getting consistently closer to 3, you might suspect that the sequence converges to 3. Convergence doesn't happen arbitrarily. It's a mathematical property that shows the sequence has some regular pattern or behavior as it "settles" around a number. Understanding convergence helps predict and understand long-term behavior of sequences in mathematics and real-life applications.
Limit Calculation Explored
Calculating the limit of a sequence helps identify the value that the terms of the sequence are approaching as \( n \) becomes very large. One common strategy to find a limit is to calculate and observe the behavior of the terms as \( n \) increases. Let's consider the sequence \( a_n = 2^n \sin(2^{-n}) \). When we compute the terms up to \( a_{10} \), we might observe that the terms seem to stabilize around 3. This suggests a limit.Mathematically, if the sequence \( a_n \) becomes arbitrarily close to a number \( L \) (which is 3 in this case), as \( n \to\infty \), then \( L \) is the limit of the sequence. This is not just a guess but a conclusion reached by examining the pattern and behavior of the terms of the sequence. Knowledge of limits is crucial in calculus, making it a fundamental tool in advanced mathematics, particularly when analyzing function behaviors and solving real-world problems.

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Most popular questions from this chapter

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n}\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k\). It can be shown that for \(n \geq 1,\) $$\left|S-\left[S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right]\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|$$ a. Interpret this inequality and explain why it gives a better approximation to \(S\) than simply using \(S_{n}\) to approximate \(S\). b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

The sequence \(\\{n !\\}\) ultimately grows faster than the sequence \(\left\\{b^{n}\right\\},\) for any \(b > 1,\) as \(n \rightarrow \infty .\) However, \(b^{n}\) is generally greater than \(n !\) for small values of \(n\). Use a calculator to determine the smallest value of \(n\) such that \(n ! > b^{n}\) for each of the cases \(b=2, b=e,\) and \(b=10\).

Consider the series \(\sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}},\) where \(p\) is a real number. a. For what values of \(p\) does this series converge? b. Which of the following series converges faster? Explain. $$ \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}} \text { or } \sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{2}} ? $$

A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=1}^{\infty} \frac{x^{k}}{k^{2}}$$
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