Problem 43 - Essential Calculus Early Transcendentals

Question

Use induction to show that the sequence defined by \(a_{1}=1\) \(a_{n+1}=3-1 / a_{n}\) is increasing and \(a_{n}<3\) for all \(n .\) Deduce that \(\left\\{a_{n}\right\\}\) is convergent and find its limit.

Step-by-Step Solution

Answer

The sequence is convergent and its limit is \( \frac{3 + \sqrt{5}}{2} \).

Step 1: Base Case

First, verify the base case for the given sequence. For \( n = 1 \), we have the given value \( a_1 = 1 \). Clearly, \( a_1 < 3 \), satisfying the second condition. Now, calculate \( a_2 = 3 - \frac{1}{a_1} = 3 - 1 = 2 \). \( a_2 > a_1 \), confirming that the sequence is increasing from the start.

Step 2: Inductive Step Assumption

Assume for an arbitrary \( n = k \) that \( a_k < 3 \) and \( a_k > a_{k-1} \). This is our induction hypothesis that we need to show holds for \( n = k+1 \).

Step 3: Inductive Step Proof

Consider \( a_{k+1} = 3 - \frac{1}{a_k} \). Since \( a_k < 3 \), then \( \frac{1}{a_k} > 0 \), so \( a_{k+1} = 3 - \frac{1}{a_k} < 3 \). Furthermore, since \( a_k > a_{k-1} \) and all terms are positive, \( \frac{1}{a_k} < \frac{1}{a_{k-1}} \), leading to \( a_{k+1} > a_k \). This shows \( a_{k+1} < 3 \) and \( a_{k+1} > a_k \), thus proving our hypothesis for \( n = k+1 \).

Step 4: Conclusion of Induction

By mathematical induction, the sequence \( \{a_n\} \) is increasing and bounded above by 3 for all \( n \). Therefore, it is confirmed that \( a_n < 3 \) and that the sequence is increasing for all terms.

Step 5: Convergence and Limit Calculation

Since the sequence is bounded and increasing, it is convergent by the Monotone Convergence Theorem. Suppose the sequence converges to limit \( L \), then both sides of the recursion \( a_{n+1} = 3 - \frac{1}{a_n} \) will approach \( L \), so \( L = 3 - \frac{1}{L} \). Solving \( L^2 = 3L - 1 \), we have \( L^2 - 3L + 1 = 0 \). Solving this quadratic equation gives roots \( L = \frac{3 \pm \sqrt{5}}{2} \). Since \( 1 < a_n < 3 \), the sequence approaches \( L = \frac{3 + \sqrt{5}}{2} \).

Key Concepts

Convergence of Sequences

A sequence is a list of numbers in a specific order. It's crucial to understand whether such a sequence converges, meaning it approaches a limiting value. If a sequence is convergent, there will be a number that it gets closer and closer to as the sequence progresses.

The sequence given in the exercise starts with a value of 1 and continues based on the formula. **Does it always move in the same direction?** Thankfully, yes! This sequence is increasing and moves towards a certain number less than 3.

To discover convergence, two main things are considered:

• **Boundedness**: Ensuring that the sequence's values do not go beyond a certain point. In our case, it's shown that the values will always be less than 3.
• **Behavior**: When you summarize the formula rules, the sequence's behavior is predictable, which strongly suggests convergence.

Since the sequence doesn't grow beyond 3 and is predicted by a regular pattern, it's determined to be convergent.

Monotone Convergence Theorem

The Monotone Convergence Theorem is an essential tool in determining whether a sequence converges. It states that every bounded sequence that is monotone—either entirely non-decreasing or non-increasing—must converge.

**Why does this matter?** In the exercise, the sequence increases but is always less than 3, making it bounded and increasing. According to the Monotone Convergence Theorem, this means that the sequence must converge to a limit.

The theorem simplifies our understanding by turning the focus on two key conditions:

• **Monotonicity**: The sequence consistently increases or decreases.
• **Boundedness**: It cannot exceed a particular threshold, in this instance, 3.

This theorem reassures us that for sequences such as the one examined here, convergence is guaranteed.

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). Solving them is often key to finding specific limits or patterns in math problems.

Within the given problem, a quadratic equation arises when setting the limit equal in the recursive formula: \( L = 3 - \frac{1}{L} \), leading to \( L^2 = 3L - 1 \). This simplifies to the quadratic equation \( L^2 - 3L + 1 = 0 \).

To solve for \( L \), factor the equation or use the quadratic formula:

• **Find the discriminant**: \( b^2 - 4ac \), which tells us whether real solutions exist.
• **Calculate solutions**: \( L_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( L = \frac{3 \pm \sqrt{5}}{2} \) is found.

This method reveals potential limits of the sequence, guiding us to the correct solution \( L = \frac{3 + \sqrt{5}}{2} \), considering the constraints \( 1 < a_n < 3 \). Understanding quadratics is vital in determining the behaviors and limits of sequences.