Question

The sequence given by $a_1 = \frac{3}{2}, a_{n+1} = 2 - \frac{1}{a_n} \quad \forall n \ge 1$ :

• A. is monotonic and converges to 2
• B. is bounded and converges to 1
• C. is bounded but not monotonic
• D. is monotonic and divergent

Hint:

For recursively defined sequences, if a limit $L$ exists, it must satisfy the relation $L = 2 - 1/L$. Solving this quadratic yields $L^2 - 2L + 1 = 0 \implies (L-1)^2 = 0 \implies L = 1$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To analyze the properties of a recursively defined sequence, we can calculate the first few terms to observe its trend and formulate an explicit expression.

Step 2: Detailed Explanation:
Let us compute the first few terms of the sequence:
- $a_1 = \frac{3}{2}$
- $a_2 = 2 - \frac{1}{a_1} = 2 - \frac{2}{3} = \frac{4}{3}$
- $a_3 = 2 - \frac{1}{a_2} = 2 - \frac{3}{4} = \frac{5}{4}$
- $a_4 = 2 - \frac{1}{a_3} = 2 - \frac{4}{5} = \frac{6}{5}$
From these terms, we can establish the general term pattern for any $n \ge 1$:
\[ a_n = \frac{n + 2}{n + 1} = 1 + \frac{1}{n + 1} \]
Let us evaluate the analytical properties of this sequence:
1. Monotonicity:
Compare $a_{n+1}$ and $a_n$:
\[ a_{n+1} - a_n = \left(1 + \frac{1}{n + 2}\right) - \left(1 + \frac{1}{n + 1}\right) = \frac{1}{n+2} - \frac{1}{n+1} \]
Since $n+2 > n+1 \implies \frac{1}{n+2} < \frac{1}{n+1}$, the difference is strictly negative:
\[ a_{n+1} - a_n < 0 \implies a_{n+1} < a_n \]
Thus, the sequence is strictly decreasing (monotonic).
2. Boundedness and Convergence:
As $n \to \infty$:
\[ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \left(1 + \frac{1}{n + 1}\right) = 1 + 0 = 1 \]
Since the sequence is strictly decreasing and bounded below by its limit 1, it is bounded:
\[ 1 < a_n \le \frac{3}{2} \]
Thus, the sequence is bounded and converges to 1.

Step 3: Final Answer
The correct option is (B).