Question

Let \(a_n\) be a sequence of real numbers such that \( a_1 = 2 \), and for \( n \geq 1 \), \( a_{n+1} = \frac{2a_n + 1}{a_n + 1} \).
 

• A. \( 1.5 \leq a_n \leq 2 \), for all natural numbers \( n \geq 1 \)
• B. There exists a natural number \( n \geq 1 \) such that \( a_n > 2 \)
• C. There exists a natural number \( n \geq 1 \) such that \( a_n < 1.5 \)
• D. There exists a natural number \( n \geq 1 \) such that \( a_n = \frac{1 + \sqrt{5}}{2} \)

Hint:

When solving recurrence relations, always check for limits and boundaries by analyzing the recursive formula and solving for the fixed points.

Answer 1

The Correct Option is A

Step 1: Analyze the recursive relation.
Given the recurrence relation, we first try to examine the behavior of the sequence. We begin with \( a_1 = 2 \). We compute the next few terms to determine if there is a pattern. Substitute \( a_1 = 2 \) into the recurrence: \[ a_2 = \frac{2(2) + 1}{2 + 1} = \frac{5}{3} \approx 1.67. \] Next, compute \( a_3 \): \[ a_3 = \frac{2(1.67) + 1}{1.67 + 1} \approx \frac{4.34}{2.67} \approx 1.63. \] Clearly, the terms are converging towards a limit.

Step 2: Solving for the limit.
To find the limit, let \( L \) be the value the sequence converges to. If the sequence converges, then \( a_{n+1} = a_n = L \). So, using the recurrence relation: \[ L = \frac{2L + 1}{L + 1}. \] Multiplying both sides by \( L + 1 \): \[ L(L + 1) = 2L + 1. \] Expanding both sides: \[ L^2 + L = 2L + 1. \] Rearranging: \[ L^2 - L - 1 = 0. \] Solving this quadratic equation: \[ L = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}. \] Thus, the two possible limits are \( L = \frac{1 + \sqrt{5}}{2} \approx 1.618 \) and \( L = \frac{1 - \sqrt{5}}{2} \). Since the sequence starts at 2 and decreases, it cannot approach the negative root, and we conclude that the sequence converges to \( \frac{1 + \sqrt{5}}{2} \). Therefore, for all \( n \geq 1 \), \( a_n \) lies between 1.5 and 2.

Step 3: Conclusion.
The correct answer is (A) \( 1.5 \leq a_n \leq 2 \), since the sequence is bounded and converges to \( \frac{1 + \sqrt{5}}{2} \), which is approximately 1.618.