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Chapter 11: Problem 5

Each of Exercises \(1-6\) gives a formula for the \(n\) th term \(a_{n}\) of a sequence \(\left\\{a_{n}\right\\} .\) Find the values of \(a_{1}, a_{2}, a_{3},\) and \(a_{4}\) . $$ a_{n}=\frac{2^{n}}{2^{n+1}} $$

Short Answer

Expert verified
All terms \(a_1, a_2, a_3, a_4\) are \(\frac{1}{2}\).

Step by step solution

01

Identify the Formula

The given formula for the sequence is\[a_{n} = \frac{2^{n}}{2^{n+1}}\]where \( n \) represents the term number. For each term, we substitute \( n \) with the desired term number.
02

Find \(a_1\)

Substitute \( n = 1 \) into the formula:\[a_1 = \frac{2^1}{2^{1+1}} = \frac{2}{4} = \frac{1}{2}\]Thus, the first term \( a_1 \) is \( \frac{1}{2} \).
03

Find \(a_2\)

Substitute \( n = 2 \) into the formula:\[a_2 = \frac{2^2}{2^{2+1}} = \frac{4}{8} = \frac{1}{2}\]Thus, the second term \( a_2 \) is \( \frac{1}{2} \).
04

Find \(a_3\)

Substitute \( n = 3 \) into the formula:\[a_3 = \frac{2^3}{2^{3+1}} = \frac{8}{16} = \frac{1}{2}\]Thus, the third term \( a_3 \) is \( \frac{1}{2} \).
05

Find \(a_4\)

Substitute \( n = 4 \) into the formula:\[a_4 = \frac{2^4}{2^{4+1}} = \frac{16}{32} = \frac{1}{2}\]Thus, the fourth term \( a_4 \) is \( \frac{1}{2} \).
06

Conclusion: Sequence Analysis

After computing the first four terms, we observe that \( a_1 = a_2 = a_3 = a_4 = \frac{1}{2} \). This indicates that the sequence is constant for these terms with the value \( \frac{1}{2} \).

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Key Concepts

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

nth term
In sequence analysis, each item in the sequence is determined by its position or term number, known as the "nth term." The nth term serves as a rule or formula that lets us determine any term in the sequence without needing to know the previous terms. This is incredibly useful, particularly for sequences with a clear pattern or rule.
For the given exercise, the nth term is represented by the formula \( a_{n} = \frac{2^{n}}{2^{n+1}} \). Here, the value of the sequence for any given term \( n \) can be found by simply substituting \( n \) into this formula.
  • \( a_1 \) is found by setting \( n = 1 \) in the formula.
  • \( a_2 \) by setting \( n = 2 \), and so on.
Understanding the nth term allows a deeper comprehension of how sequences evolve, providing a precise way to analyze patterns within sequences.
constant sequence
When we talk about sequences, a "constant sequence" is one where all terms are the same. This means that no matter which term number you look at, it always has the same value. In the context of our exercise, we encountered a constant sequence.
The sequential evaluation shows that the terms \( a_1 = a_2 = a_3 = a_4 \) all equal \( \frac{1}{2} \). This constant value indicates that regardless of the term number we've calculated up to, the output remains unchanged.
  • Constant sequences are predictable as they do not vary over time.
  • They provide a great example of stability within mathematical functions.
They are easy to analyze since every term is the same, but understanding them can reveal interesting insights about the nature of the formula creating them. Even though each term remains identical, exploring why it stays constant is key in deeper sequence analysis.
formula substitution
Formula substitution is a core technique for solving sequence problems. It involves replacing a variable, like \( n \) in our nth term formula, with specific numbers to find the values of sequence terms. This straightforward method allows us to gain concrete values from a general formula.
In this exercise, you repeatedly replace \( n \) in the formula \(\frac{2^{n}}{2^{n+1}}\) with individual term numbers like 1, 2, 3, and 4, to determine \( a_1, a_2, a_3, \) and \( a_4 \).
  • This step-by-step replacement gives tangible terms from an abstract formula.
  • It provides a clear method to verify and compute each sequence's term.
By simply performing these substitutions, any sequence value can be conveniently reached, making it an essential tool for anyone looking to master sequence analysis.

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Most popular questions from this chapter

Logistic difference equation The recursive relation $$ a_{n+1}=r a_{n}\left(1-a_{n}\right) $$ is called the logistic difference equation, and when the initial value \(a_{0}\) is given the equation defines the logistic sequence \(\left\\{a_{n}\right\\} .\) Throughout this exercise we choose \(a_{0}\) in the interval \(03.57\) . Choose \(r=3.65\) and calculate and plot the first 300 terms of \(\left\\{a_{n}\right\\} .\) Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of \(a_{n+1}\) from previous values of the sequence. g. For \(r=3.65\) choose two starting values of \(a_{0}\) that are close together, say, \(a_{0}=0.3\) and \(a_{0}=0.301 .\) Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for \(r=3.75 .\) Can you see how the plots look different depending on your choice of \(a_{0} ?\) We say that the logistic sequence is sensitive to the initial condition a_{0} .

Which of the sequences in Exercises \(101-106\) converge, and which diverge? Give reasons for your answers. $$ a_{n}=\left((-1)^{n}+1\right)\left(\frac{n+1}{n}\right) $$

Outline of the proof of the Rearrangement Theorem a. Let \(\epsilon\) be a positive real number, let \(L=\sum_{n=1}^{\infty} a_{n}\), and let \(s_{k}=\sum_{n=1}^{k} a_{n} .\) Show that for some index \(N_{1}\) and for some index \(N_{2} \geq N_{1}\), $$ \sum_{n=N_{1}}^{\infty}\left|a_{n}\right|<\frac{\epsilon}{2} \text { and }\left|s_{N_{2}}-L\right|<\frac{\epsilon}{2} $$ Since all the terms \(a_{1}, a_{2}, \ldots, a_{N_{2}}\) appear somewhere in the sequence \(\left\\{b_{n}\right\\}\), there is an index \(N_{3} \geq N_{2}\) such that if \(n \geq N_{3}\), then \(\left(\sum_{k=1}^{n} b_{k}\right)-s_{N_{2}}\) is at most a sum of terms \(a_{m}\) with \(m \geq N_{1}\). Therefore, if \(n \geq N_{3}\), $$ \begin{aligned} \left|\sum_{k=1}^{n} b_{k}-L\right| & \leq\left|\sum_{k=1}^{n} b_{k}-s_{N_{2}}\right|+\left|s_{N_{2}}-L\right| \\ & \leq \sum_{k=N_{1}}^{\infty}\left|a_{k}\right|+\left|s_{N_{2}}-L\right|<\epsilon \end{aligned} $$ b. The argument in part (a) shows that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely then \(\sum_{n=1}^{\infty} b_{n}\) converges and \(\sum_{n=1}^{\infty} b_{n}=\sum_{n=1}^{\infty} a_{n}\) Now show that because \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) converges, \(\sum_{n=1}^{\infty}\left|b_{n}\right|\) converges to \(\sum_{n=1}^{\infty}\left|a_{n}\right|\).

a. Suppose that \(f(x)\) is differentiable for all \(x\) in \([0,1]\) and that \(f(0)=0 .\) Define the sequence \(\left\\{a_{n}\right\\}\) by the rule \(a_{n}=\) \(n f(1 / n) .\) Show that \(\lim _{n \rightarrow \infty} a_{n}=f^{\prime}(0)\) Use the result in part (a) to find the limits of the following sequences \(\left\\{a_{n}\right\\}\). \(\begin{array}{ll}{\text { b. } a_{n}=n \tan ^{-1} \frac{1}{n}} & {\text { c. } a_{n}=n\left(e^{1 / n}-1\right)} \\ {\text { d. } a_{n}=n \ln \left(1+\frac{2}{n}\right)}\end{array}\)

The sequence \(\\{n /(n+1)\\}\) has a least upper bound of 1 Show that if \(M\) is a number less than \(1,\) then the terms of \(\\{n /(n+1)\\}\) eventually exceed \(M .\) That is, if \(M<1\) there is an integer \(N\) such that \(n /(n+1)>M\) whenever \(n>N .\) since \(n /(n+1)<1\) for every \(n,\) this proves that 1 is a least upper bound for \(\\{n /(n+1)\\}\).
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