Problem 81 - Calculus: Early Transcendentals

Question

Show that the sequence defined by \( a_1 = 1 \) \( a_{n + 1} = 3 - \frac{1}{a_n} \) is increasing and \( a_n < 3 \) for all \( n. \) Deduce that \( \\{ a_n \\} \) is convergent and find its limit.

Step-by-Step Solution

Answer

The sequence is increasing, bounded by 3, convergent, with limit \( \frac{3-\sqrt{5}}{2} \).

Step 1: Define the sequence properties

The sequence is given by \( a_1 = 1 \) and \( a_{n+1} = 3 - \frac{1}{a_n} \). We need to show that this sequence is increasing and that each term is less than 3.

Step 2: Base case verification for boundedness

First, we verify that the sequence is initially bounded by 3. We have \( a_1 = 1 < 3 \). Assume \( a_n < 3 \), we need to prove \( a_{n+1} < 3 \).

Step 3: Inductive step for boundedness

From the formula \( a_{n+1} = 3 - \frac{1}{a_n} \), we know that \( a_{n+1} < 3 \) since \( \frac{1}{a_n} > 0 \). This shows if \( a_n < 3 \), then \( a_{n+1} < 3 \). Induction confirms boundedness by 3.

Step 4: Verify increasing nature

We need to show that \( a_{n+1} > a_n \). Rewriting the inequality \( 3 - \frac{1}{a_n} > a_n \) leads to \( 3 > a_n + \frac{1}{a_n} \). Multiplying both sides by \( a_n \), we get \( 3a_n > a_n^2 + 1 \), or \( 0 > (a_n-1)(a_n-2) \). Since \( a_n < 3 \), we see the inequality holds for \( 1 < a_n < 3 \). This proves the sequence is increasing.

Step 5: Sequence convergence and limit

The sequence is increasing and bounded by 3. Therefore, it converges. Let the limit be \( L \), then substituting in the recurrence relation, \( L = 3 - \frac{1}{L} \). Solving \( L^2 = 3L - 1 \), we find \( L^2 - 3L + 1 = 0 \). The solutions are \( L = \frac{3 \pm \sqrt{5}}{2} \). Since \( a_n < 3 \), the positive solution, \( L = \frac{3 - \sqrt{5}}{2} \), is the limit.

Key Concepts

Mathematical Induction

Mathematical induction is a proof technique used to show that a statement holds true for all natural numbers. It involves two steps: the base case and the inductive step. For the given sequence, induction helps us establish that all terms are less than 3 by confirming:

• The base case: Verify the initial condition, which is that the first term, \( a_1 = 1 \), is indeed less than 3.
• Inductive Step: Assume the statement is true for an arbitrary term \( a_n \), i.e., \( a_n < 3 \), and then prove it is true for \( a_{n+1} \). Our step shows that if \( a_n < 3 \), the next term \( a_{n+1} = 3 - \frac{1}{a_n} \) also satisfies this condition, because \( \frac{1}{a_n} > 0 \).

By following these steps, the entire sequence \( \{ a_n \} \) is shown to be bounded above by 3.

Bounded Sequence

A bounded sequence is one that has both an upper and a lower bound. For the problem at hand, the sequence is confirmed to be bounded above by 3.

• Starting with \( a_1 = 1 \), we show this term is below 3.
• The recursive formula ensures that every subsequent term is computed using: \( a_{n+1} = 3 - \frac{1}{a_n} \), hence remains less than 3 since \( \frac{1}{a_n} \) is positive.
• Inductive reasoning confirms that for every \( n \), \( a_n < 3 \); thus, the sequence cannot exceed this upper limit.

Knowing a sequence's bounds helps ascertain its behavior over time—essential information when proving convergence.

Increasing Sequence

An increasing sequence is a sequence where each term is greater than or equal to the one preceding it. To show that \( \{ a_n \} \) is increasing, we prove:

• The formula \( a_{n+1} = 3 - \frac{1}{a_n} \) gives rise to the inequality \( a_{n+1} > a_n \) or equivalently, \( 3 - \frac{1}{a_n} > a_n \).
• Manipulating this inequality leads to \( 3 > a_n + \frac{1}{a_n} \), which can be rearranged to show that \( 0 > (a_n-1)(a_n-2) \).
• Since we have established \( 1 < a_n < 3 \) from boundedness, the inequality holds. Therefore, as \( a_n \) increases, each subsequent term still fits within the range, validating the sequence's increasing nature.

This characteristic in combination with boundedness is crucial for confirming convergence.

Limit of a Sequence

The limit of a sequence is the value that the terms approach as the sequence progresses. For \( \{ a_n \} \), we've ascertained it’s both increasing and bounded, hence it must converge.

• Assuming the limit \( L \) exists, substitute into the recurrence: \( L = 3 - \frac{1}{L} \).
• Solving \( L^2 = 3L - 1 \) leads to the quadratic equation \( L^2 - 3L + 1 = 0 \).
• The solutions of this equation are \( L = \frac{3 \pm \sqrt{5}}{2} \), which are derived using the quadratic formula.
• Given the constraint \( a_n < 3 \), choose the solution \( L = \frac{3 - \sqrt{5}}{2} \) as it fits the sequence's bounds.

The limit confirms how the sequence behaves over the long run, providing insight into its final state.