Problem 36 - Essential Calculus Early Transcendentals

Question

(a) If \(\left\\{a_{n}\right\\}\) is convergent, show that $$\lim _{n \rightarrow \infty} a_{n+1}=\lim _{n \rightarrow \infty} a_{n}$$ (b) A sequence \(\left\\{a_{n}\right\\}\) is defined by \(a_{1}=1\) and \(a_{n+1}=1 /\left(1+a_{n}\right)\) for \(n \geqslant 1 .\) Assuming that \(\left\\{a_{n}\right\\}\) is convergent, find its limit.

Step-by-Step Solution

Answer

The limit is \(\frac{-1 + \sqrt{5}}{2}\).

Step 1: Define the Limit of Convergence

Let \( L \) be the limit of the sequence \( \{a_n\} \), so \( \lim_{n \to \infty} a_n = L \). Since \( \{a_n\} \) is convergent, it follows that \( \lim_{n \to \infty} a_{n+1} = L \) as well. This satisfies the condition \( \lim_{n \rightarrow \infty} a_{n+1} = \lim_{n \rightarrow \infty} a_{n} = L \). The sequence values do not depend on \( n \), only the limit matters.

Step 2: Set Up the Recursive Relation

We know the recursive formula given by \( a_{n+1} = \frac{1}{1 + a_n} \). Assuming the sequence \( \{a_n\} \) is convergent, substitute the limit \( L \) into this recursive formula: \[ L = \frac{1}{1 + L} \].

Step 3: Solve the Limit Equation

Starting from the equation \( L = \frac{1}{1 + L} \), multiply both sides by \(1 + L\): \[ L(1 + L) = 1 \].This gives \[ L^2 + L = 1 \].Rearrange to form a quadratic equation: \[ L^2 + L - 1 = 0 \].

Step 4: Solve the Quadratic Equation

Use the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for \( a = 1, b = 1, c = -1 \):\[ L = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \].

Step 5: Select the Meaningful Root

Since limits of a sequence cannot be negative, select the positive root:\( L = \frac{-1 + \sqrt{5}}{2} \).

Step 6: Conclusion

The limit of the sequence \( \{a_n\} \) is \( \lim_{n \to \infty} a_n = \frac{-1 + \sqrt{5}}{2} \).

Key Concepts

Limit of a Sequence

In mathematics, a sequence is simply a list of numbers ordered in a specific way. A sequence can converge, which means it approaches a specific value, called the limit, as it progresses. When we say that a sequence \( \{a_n\} \) is convergent, we mean that as \( n \) (an index in the sequence) goes to infinity, \( a_n \) gets closer and closer to a fixed number \( L \).
This concept is central in calculus and analysis.
A convergent sequence has a unique limit, and all elements beyond a certain point in the sequence fall arbitrarily close to this limit. For example, if \( L \) is the limit, no matter how small the distance you choose, there will be a point in the sequence where all further terms are closer than that distance to \( L \).

• Key property: \( \lim_{n \rightarrow \infty} a_n = L \) implies \( \lim_{n \rightarrow \infty} a_{n+1} = L \).
• Independence: The convergence is not about the sequence's terms changing when \( n \) is larger, but about how the terms behave as a whole.

Recursive Sequences

A recursive sequence is one in which future elements are defined based on preceding elements. Often, a formula or rule dictates how to calculate the next term using the terms before it. This recursive formula captures the essence of the sequence.
For example, the sequence we are considering is defined by the recursive relation \( a_{n+1} = \frac{1}{1 + a_n} \). Here, each term in the sequence relies solely on the value of the term that comes right before it.
These sequences might converge to a limit.

• Initial value: Typically, the first term \( a_1 \) is given (such as \( a_1 = 1 \) in our example).
• Recursive formula: The challenge often involves using this formula to uncover the limit of the sequence.

Quadratic Equations

Quadratic equations appear frequently, especially when solving problems involving limits and recursive sequences. A quadratic equation has the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.
The solutions to a quadratic equation can be found using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In solving our limit problem, we transformed the recursive limit equation into a quadratic equation: \( L^2 + L - 1 = 0 \). By solving this equation using the quadratic formula, we derived the two possible values for \( L \). Given constraints, such as positivity in our limit scenario, helped refine which solution (root) was meaningful. The selected root was \( L = \frac{-1 + \sqrt{5}}{2} \), demonstrating how quadratic equations can provide critical insights in recursive sequence analysis.

• Discriminant (\( b^2 - 4ac \)) informs the nature of the roots.
• Choosing the correct root can depend on the real-world context or additional constraints.