Question

For the sequence \((x_n)\), if \(x_1 > 1\) and \(x_{n+1} = 2 - \frac{1}{x_n}\), \(n \geq 1\), which of the following is true:

• A. \((x_n)\) is monotonically increasing
• B. \((x_n)\) is monotonically decreasing
• C. \((x_n)\) is oscillating
• D. \((x_n)\) is unbounded

Hint:

Exam Tip:
For recurrence relations:

• Find fixed points.
• Check monotonicity by comparing \(x_{n+1}\) and \(x_n\).
• The sequence is often monotonic if the function is continuous and the initial value is on one side of the fixed point.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We have a recurrence relation \(x_{n+1} = 2 - \frac{1}{x_n}\) with \(x_1 > 1\). We need to determine the behavior of the sequence.

Step 2: Key Formula or Approach:
Let's find the fixed point: \(L = 2 - \frac{1}{L} \Rightarrow L^2 - 2L + 1 = 0 \Rightarrow (L - 1)^2 = 0 \Rightarrow L = 1\).
So, the sequence may converge to 1.

Step 3: Detailed Explanation:
Let's compute a few terms for \(x_1 = 2\): \[ x_2 = 2 - \frac{1}{2} = \frac{3}{2} = 1.5 \] \[ x_3 = 2 - \frac{1}{1.5} = 2 - \frac{2}{3} = \frac{4}{3} \approx 1.333 \] \[ x_4 = 2 - \frac{1}{1.333} = 2 - 0.75 = 1.25 \] The sequence is decreasing: 2, 1.5, 1.333, 1.25, \ldots
It is monotonically decreasing and bounded below by 1.
So, option (B) is correct.
Let's check if the sequence is increasing for other initial values.
If \(x_1 > 1\), then \(x_2 = 2 - \frac{1}{x_1}\).
We want to check if \(x_2 > x_1\):
\(2 - \frac{1}{x_1} > x_1 \Rightarrow 2 - x_1 > \frac{1}{x_1} \Rightarrow x_1(2 - x_1) > 1 \Rightarrow -x_1^2 + 2x_1 - 1 > 0 \Rightarrow -(x_1 - 1)^2 > 0\), which is never true for \(x_1 \neq 1\).
So, \(x_2 < x_1\).
Thus, the sequence is decreasing.
So, option (B) is correct.

Step 4: Final Answer:
Therefore, option (B) is correct.