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Chapter 9: Problem 47

Consider the following sequences. a. Find the first four terms of the sequence. b. Based on part (a) and the figure, determine a plausible limit of the sequence. $$a_{n}=2+2^{-n} ; n=1,2,3, \ldots$$

Short Answer

Expert verified
Answer: The plausible limit of the sequence is 2.

Step by step solution

01

Find the first four terms of the sequence

To find the first four terms of the sequence, simply substitute the values of n in the given formula: $$a_{n}=2+2^{-n}$$ For n=1: $$a_{1}=2+2^{-1}=2+\frac{1}{2}=\frac{5}{2}$$ For n=2: $$a_{2}=2+2^{-2}=2+\frac{1}{4}=\frac{9}{4}$$ For n=3: $$a_{3}=2+2^{-3}=2+\frac{1}{8}=\frac{17}{8}$$ For n=4: $$a_{4}=2+2^{-4}=2+\frac{1}{16}=\frac{33}{16}$$ So, the first 4 terms of the sequence are: $$\frac{5}{2}, \frac{9}{4}, \frac{17}{8}, \frac{33}{16}$$
02

Determine a plausible limit for the sequence

Now that we have the first 4 terms of the sequence, let's try to analyze their behavior to find a plausible limit. Looking at the terms, we notice that they are increasing, but the rate at which they increase is decreasing as n gets larger. This is due to the 2^{-n} term in the sequence. As n gets larger, 2^{-n} gets smaller and approaches to zero. The plausible limit can be obtained by considering what happens to the expression as n approaches infinity: $$\lim_{n\to\infty} a_n = \lim_{n\to\infty} (2+2^{-n})$$ Since the term $$2^{-n}$$ tends to 0 as n approaches infinity: $$\lim_{n\to\infty} (2+2^{-n})=2+0=2$$ Thus, it seems reasonable to assume that the plausible limit of the given sequence is 2.

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Key Concepts

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

Limit of a Sequence
A sequence is a list of numbers where you can always calculate the next number using a specific formula. The **limit** of a sequence is the value that the numbers in the sequence get closer to as you keep calculating more and more terms. It is like the "goal" of the sequence. In our example, the sequence given is \( a_n = 2 + 2^{-n} \).
  • As \( n \) gets larger, the term \( 2^{-n} \) becomes very small. Imagine \( 2^{-10} \), which is \( \frac{1}{1024} \), a tiny number!
  • Therefore, \( a_n \) increasingly looks like \( 2 + 0 \), which is just 2.
  • A sequence "has a limit" if it eventually gets as close as you want to the limit. We found that for our sequence (as shown in the solution), the plausible limit is 2.
Knowing the limit helps us understand the long-term behavior of a sequence.
Convergent Sequence
A sequence is called **convergent** when its terms get closer and closer to a specific limit as you move further out in the sequence. This means that for any small number you can think of, you can find a point in the sequence after which the terms are all within that small distance of the limit.In the example with \( a_n = 2 + 2^{-n} \):
  • As \( n \) becomes larger, the \( 2^{-n} \) part shrinks toward zero.
  • Consequently, the whole sequence \( a_n \) gradually approaches the number 2, its limit.
  • Because \( a_n \) is consistently moving closer to its limit and doesn't diverge elsewhere, we can say it's a convergent sequence.
This predictability in a sequence's behavior is what makes working with convergent sequences so valuable in mathematics.
Infinite Series
An **infinite series** is essentially the sum of the terms of an infinite sequence. If you keep adding the terms, you might arrive at a total that is also converging to a limit.Here's how it connects to sequences:
  • An infinite series is often denoted as \( S = a_1 + a_2 + a_3 + \ldots \).
  • If the sequence \( a_n \) converges to a limit, it's possible that the series made from its terms can also have a specific sum that it approaches.
  • The behavior and convergence of infinite series are central to many areas of calculus and mathematical analysis.
In many mathematical situations, especially in calculus, determining whether an infinite series converges can tell you a lot about the function or system you are studying.

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Most popular questions from this chapter

Infinite products An infinite product \(P=a_{1} a_{2} a_{3} \ldots,\) which is denoted \(\prod_{k=1}^{\infty} a_{k}\) is the limit of the sequence of partial products \(\left\\{a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \dots\right\\}\) a. Show that the infinite product converges (which means its sequence of partial products converges) provided the series \(\sum_{k=1}^{\infty} \ln a_{k}\) converges. b. Consider the infinite product $$P=\prod_{k=2}^{\infty}\left(1-\frac{1}{k^{2}}\right)=\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} \cdot \frac{24}{25} \cdots$$ Write out the first few terms of the sequence of partial products, $$P_{n}=\prod_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right)$$ (for example, \(P_{2}=\frac{3}{4}, P_{3}=\frac{2}{3}\) ). Write out enough terms to determine the value of the product, which is \(\lim _{n \rightarrow \infty} P_{n}\). c. Use the results of parts (a) and (b) to evaluate the series $$\sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right)$$

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The CORDIC (COordinate Rotation DIgital Calculation) algorithm is used by most calculators to evaluate trigonometric and logarithmic functions. An important number in the CORDIC algorithm, called the aggregate constant, is \(\prod_{n=0}^{\infty} \frac{2^{n}}{\sqrt{1+2^{2 n}}},\) where \(\prod_{n=0}^{N} a_{n}\) represents the product \(a_{0} \cdot a_{1} \cdots a_{N}\). This infinite product is the limit of the sequence $$\left\\{\prod_{n=0}^{0} \frac{2^{n}}{\sqrt{1+2^{2 n}}} \cdot \prod_{n=0}^{1} \frac{2^{n}}{\sqrt{1+2^{2 n}}}, \prod_{n=0}^{2} \frac{2^{n}}{\sqrt{1+2^{2 n}}} \ldots .\right\\}.$$ Estimate the value of the aggregate constant.
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