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Chapter 8: Problem 13

Write the first four terms of the sequence \(\left\\{a_{n}\right\\}_{n=1}^{\infty}\) $$a_{n}=\frac{2^{n+1}}{2^{n}+1}$$

Short Answer

Expert verified
Answer: The first four terms of the sequence are \(\frac{4}{3}, \frac{8}{5}, \frac{16}{9}, \frac{32}{17}\).

Step by step solution

01

Calculate \(a_1\)

To find \(a_{1}\), plug in n=1 into the formula: \(a_{1}=\frac{2^{1+1}}{2^{1}+1}\). Now simplify: \(a_{1}=\frac{2^{2}}{2^{1}+1}=\frac{4}{2+1}=\frac{4}{3}\).
02

Calculate \(a_2\)

To find \(a_{2}\), plug in n=2 into the formula: \(a_{2}=\frac{2^{2+1}}{2^{2}+1}\). Now simplify: \(a_{2}=\frac{2^{3}}{2^{2}+1}=\frac{8}{4+1}=\frac{8}{5}\).
03

Calculate \(a_3\)

To find \(a_{3}\), plug in n=3 into the formula: \(a_{3}=\frac{2^{3+1}}{2^{3}+1}\). Now simplify: \(a_{3}=\frac{2^{4}}{2^{3}+1}=\frac{16}{8+1}=\frac{16}{9}\).
04

Calculate \(a_4\)

To find \(a_{4}\), plug in n=4 into the formula: \(a_{4}=\frac{2^{4+1}}{2^{4}+1}\). Now simplify: \(a_{4}=\frac{2^{5}}{2^{4}+1}=\frac{32}{16+1}=\frac{32}{17}\).
05

Write down the first four terms

Now that we have calculated the first four terms, we can write them down as: $$a_{1}=\frac{4}{3}, \ a_{2}=\frac{8}{5}, \ a_{3}=\frac{16}{9}, \ a_{4}=\frac{32}{17}$$ These are the first four terms of the sequence \(\left\\{a_{n}\right\\}_{n=1}^{\infty}\), defined by the formula \(a_{n}=\frac{2^{n+1}}{2^{n}+1}\).

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Key Concepts

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

Recurrence Relations
A recurrence relation is a way of defining a sequence where each term is specified as a function of its preceding terms. Without having to calculate from scratch, these relations allow us to generate terms efficiently one by one. For example, in a sequence defined by a recurrence relation such as \( a_n = a_{n-1} + 2 \), each term depends directly on the previous term. This recursive formula can be managed and solved by repeatedly applying the function to the starting condition or known terms. Recurrence relations require an initial condition to begin deriving terms of the sequence, often known as the "seed" or "initial term." Typically, this type of sequence is useful in computing scenarios where iterative processes are prevalent, such as algorithms and programming.
Sequence Convergence
Sequence convergence occurs when the terms of a sequence progressively get closer and closer to a certain fixed value as the sequence progresses to infinity. We call this fixed value the "limit" of the sequence. Mathematically, a sequence \( \{a_n\} \) converges to a limit \( L \) if, for any small number \( \epsilon > 0 \), there exists a natural number \( N \) such that for all \( n > N \), it holds true that \( |a_n - L| < \epsilon \). When examining a sequence for convergence, it's crucial to assess how the behavior of its terms changes as \( n \) becomes larger. A sequence may also diverge, which means it does not approach a single finite limit, and understanding this distinction is vital when working with infinite sequences.
Fraction Simplification
Simplifying fractions is the process of reducing the numerator and denominator to their smallest possible values while maintaining the same ratio. This frequently involves dividing both the top (numerator) and bottom (denominator) by their greatest common divisor (GCD). Simplification enhances clarity, making fractions easier to understand and work with. For example, the fraction \( \frac{8}{12} \) can be simplified by dividing both the top and bottom by their GCD, which is 4, resulting in \( \frac{2}{3} \). In sequences, particularly those defined in fractional form like our example sequence \( a_n = \frac{2^{n+1}}{2^n + 1} \), fraction simplification is crucial in ensuring we work with the terms in their most efficient state. Recognizing common factors and applying arithmetic operations efficiently can simplify complex fractional sequences.

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Most popular questions from this chapter

Stirling's formula Complete the following steps to find the values of \(p>0\) for which the series \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges. a. Use the Ratio Test to show that \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges for \(p>2\). b. Use Stirling's formula, \(k !=\sqrt{2 \pi k} k^{k} e^{-k}\) for large \(k,\) to determine whether the series converges when \(p=2\). (Hint: \(1 \cdot 3 \cdot 5 \cdots(2 k-1)=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots(2 k-1) 2 k}{2 \cdot 4 \cdot 6 \cdots 2 k}\) (See the Guided Project Stirling's formula and \(n\) ? for more on this topic.)

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{n}{n^{2}+1}=0$$

Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \dots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots$$

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(\left|R_{n}\right|<10^{-6}\) ). Functions defined as series Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} x^{k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. The Consumer Price Index (the CPI is a measure of the U.S. cost of living) is given a base value of 100 in the year \(1984 .\) Assume the CPI has increased by an average of \(3 \%\) per year since \(1984 .\) Let \(c_{n}\) be the CPI \(n\) years after \(1984,\) where \(c_{0}=100\)
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