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Chapter 9: Problem 102

Assume that each sequence converges and find its limit. $$a_{1}=-1, \quad a_{n+1}=\frac{a_{n}+6}{a_{n}+2}$$

Short Answer

Expert verified
The limit of the sequence is 2.

Step by step solution

01

Understand the sequence definition

The sequence is defined recursively, with \(a_1 = -1\) and each subsequent term \(a_{n+1}\) is given by the formula \(a_{n+1} = \frac{a_n + 6}{a_n + 2}\). We are tasked with finding the limit of this sequence as \(n\) approaches infinity.
02

Assume the limit exists

Let's assume the sequence \((a_n)\) converges to a limit \(L\). By the properties of limits, this means that both \(a_n\) and \(a_{n+1}\) approach \(L\) as \(n\) approaches infinity.
03

Set the recursive formula equal to the limit

Since \(a_{n+1} = \frac{a_n + 6}{a_n + 2}\) and both \(a_n\) and \(a_{n+1}\) converge to \(L\), we can set:\[ L = \frac{L + 6}{L + 2} \]
04

Solve for the limit \(L\)

To solve \(L = \frac{L + 6}{L + 2}\), cross-multiply to get:\[ L(L + 2) = L + 6 \]Expand this to get:\[ L^2 + 2L = L + 6 \]Rearrange this into a standard quadratic form:\[ L^2 + L - 6 = 0 \]
05

Solve the quadratic equation

Use the quadratic formula \(L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve \(L^2 + L - 6 = 0\), where \(a = 1\), \(b = 1\), and \(c = -6\):\[ L = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \]\[ L = \frac{-1 \pm \sqrt{1 + 24}}{2} \]\[ L = \frac{-1 \pm \sqrt{25}}{2} \]\[ L = \frac{-1 \pm 5}{2} \]Thus, \(L = 2\) or \(L = -3\).
06

Determine the valid limit based on initial conditions

To determine which solution is valid, examine the sequence's possible behavior. Start at \(a_1 = -1\). Since \(a_n\) might alternate based on the recursive formula, check if the sequence approaches either of the limits. Usually, a negative start with the given formula gravitates to positive convergence if no alternating behavior negates it. Only \(L = 2\) is possible under a consistent positive recursive outcome.

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Key Concepts

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

Limit of a Sequence
The concept of the limit of a sequence is fundamental in calculus and analysis. A sequence is a list of numbers arranged in a specific order, and when we discuss its limit, we are interested in what value, if any, the terms of the sequence approach as we progress towards infinity.
When a sequence converges, its terms get closer and closer to a specific number known as the limit. For example, in our problem, we are tasked with finding the limit of the sequence \(a_n\) as the term number \(n\) goes to infinity. We assume that a limit \(L\) exists for this sequence.
  • Convergence: A sequence \(a_n\) is said to converge to a limit \(L\) if for any small number \(\varepsilon > 0\), there exists a number \(N\) such that for all \(n > N\), the absolute difference \(|a_n - L|\) is less than \(\varepsilon\).

  • Divergence: If such a limit does not exist, the sequence is said to diverge.
Calculating the limit is crucial for understanding the behavior of sequences in mathematics, particularly when solving practical problems involving recursive sequences or other advanced mathematical functions.
Recursive Sequence
A recursive sequence is a sequence in which each term after the first one is defined as a function of the preceding terms. In our exercise, the sequence \(a_n\) is defined by the initial term \(a_1 = -1\), and each subsequent term \(a_{n+1}\) is calculated using the formula \(a_{n+1} = \frac{a_n + 6}{a_n + 2}\).

  • Initial Term: This is the starting point of the sequence, which in our case is \(a_1 = -1\).

  • Recursive Formula: This formula \(a_{n+1} = \frac{a_n + 6}{a_n + 2}\) is applied repeatedly to find the terms of the sequence.
Understanding recursive sequences involves recognizing patterns and applying the recursive formula until a clear understanding of the sequence's behavior is obtained. For sequences that converge, recursive definitions help in establishing limits and understanding the sequence's long-term behavior.
Quadratic Equation Solution
Solving quadratic equations is a common mathematical task, essential for finding the roots of equations that have specific forms, such as \(ax^2 + bx + c = 0\).
In our step-by-step solution, we reached the equation \(L^2 + L - 6 = 0\) while trying to find the sequence's limit when \(L\) is assumed to be the converging value.

  • Quadratic Formula: The solution to a quadratic equation \(ax^2 + bx + c = 0\) is given by \(L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

  • Applying the formula, we find \(L = \frac{-1 \pm \sqrt{25}}{2}\), resulting in the possible limits \(L = 2\) or \(L = -3\).
In our situation, further analysis of the sequence's behavior helped us choose \(L = 2\) as the valid limit, based on the initial term and the conditions given. Understanding how to solve quadratic equations and relate the roots back to the context, like in sequence convergence, provides valuable insights in various mathematical scenarios.

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Most popular questions from this chapter

Series for \(\tan ^{-1} x\) for \(|x|>1\) Derive the series $$\begin{array}{l}\tan ^{-1} x=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x>1 \\\\\tan ^{-1} x=-\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x<-1\end{array}$$ by integrating the series $$\frac{1}{1+t^{2}}=\frac{1}{t^{2}} \cdot \frac{1}{1+\left(1 / t^{2}\right)}=\frac{1}{t^{2}}-\frac{1}{t^{4}}+\frac{1}{t^{6}}-\frac{1}{t^{8}}+\cdots$$ in the first case from \(x\) to \(\infty\) and in the second case from \(-\infty\) to \(x\).

The series $$\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9 !}-\frac{x^{11}}{11 !}+\cdots$$ converges to \(\sin x\) for all \(x\). a. Find the first six terms of a series for \(\cos x .\) For what values of \(x\) should the series converge? b. By replacing \(x\) by \(2 x\) in the series for \(\sin x,\) find a series that converges to \(\sin 2 x\) for all \(x\) c. Using the result in part (a) and series multiplication, calculate the first six terms of a series for \(2 \sin x \cos x .\) Compare your answer with the answer in part (b).

Assume that \(b_{n}\) is a sequence of positive numbers converging to 1/3. Determine whether the following series converge or diverge. a. \(\sum_{n=1}^{\infty} \frac{b_{n+1} b_{n}}{n 4^{n}}\) b. \(\sum_{n=1}^{\infty} \frac{n^{n}}{n ! b_{1}^{2} b_{2}^{2} \cdots b_{n}^{2}}\)

Which of the series \(\Sigma_{n=1}^{\infty} a_{n}\) defined by the formulas converge, and which diverge? Give reasons for your answers. $$a_{1}=1, \quad a_{n+1}=\frac{1+\ln n}{n} a_{n}$$

Which of the series \(\Sigma_{n=1}^{\infty} a_{n}\) defined by the formulas converge, and which diverge? Give reasons for your answers. $$a_{1}=3, \quad a_{n+1}=\frac{n}{n+1} a_{n}$$
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