Question

Let \(\{a_n\}_{n\ge1}\) and \(\{b_n\}_{n\ge1}\) be two convergent sequences of real numbers. For \( n \geq 1 \), define \( u_n = \max\{a_n, b_n\} \) and \( v_n = \min\{a_n, b_n\} \). Then
 

• A. Neither \(\{u_n\}_{n\ge1}\) nor \(\{v_n\}_{n\ge1}\) converges
• B. \(\{u_n\}_{n\ge1}\) converges but \(\{v_n\}_{n\ge1}\) does not converge
• C. \(\{u_n\}_{n\ge1}\) does not converge but \(\{v_n\}_{n\ge1}\) converges
• D. Both \(\{u_n\}_{n\ge1}\) and \(\{v_n\}_{n\ge1}\) converge

Hint:

When dealing with the maximum and minimum of convergent sequences, remember that these operations preserve convergence and the limits are simply the maximum or minimum of the individual sequence limits.

Answer 1

The Correct Option is D

Step 1: Understanding convergence of sequences. 
Since \( a_n \) and \( b_n \) are both convergent sequences, let: \[ \lim_{n \to \infty} a_n = A \text{and} \lim_{n \to \infty} b_n = B. \] By the properties of limits, the maximum and minimum of two convergent sequences are also convergent. Specifically: \[ \lim_{n \to \infty} u_n = \lim_{n \to \infty} \max(a_n, b_n) = \max(A, B), \] \[ \lim_{n \to \infty} v_n = \lim_{n \to \infty} \min(a_n, b_n) = \min(A, B). \]

Step 2: Analyzing the options. 
- (A) Neither \(\{u_n\}_{n\ge1}\) nor \(\{v_n\}_{n\ge1}\) converges: This is incorrect, as both sequences are given as convergent. 
- (B) \(\{u_n\}_{n\ge1}\) converges but \(\{v_n\}_{n\ge1}\) does not converge: This is incorrect, as both sequences \( u_n \) and \( v_n \) converge to \( \max(A, B) \) and \( \min(A, B) \), respectively. 
- (C)\(\{u_n\}_{n\ge1}\) does not converge but \(\{v_n\}_{n\ge1}\) converges: This is incorrect for the same reason as (B). 
- (D) Both \(\{u_n\}_{n\ge1}\) and \(\{v_n\}_{n\ge1}\) converge: This is the correct answer, as both sequences converge by the properties of limits.

Step 3: Conclusion. 
The correct answer is (D) Both \(\{u_n\}_{n\ge1}\) and \(\{v_n\}_{n\ge1}\) converge, as both the maximum and minimum of two convergent sequences are convergent.