Question

For \( n \in \mathbb{N} \), define \( x_n \) and \( y_n \) by \[ x_n = (-1)^n \frac{3n}{n^3} \quad \text{and} \quad y_n = \left(4n^3 + (-1)^n 3n^3 \right)^{1/n}. \] Then, which one of the following is TRUE?

• A. \( (x_n) \) has a convergent subsequence, and NO subsequence of \( (y_n) \) is convergent.
• B. NO subsequence of \( (x_n) \) is convergent, and \( (y_n) \) has a convergent subsequence.
• C. \( (x_n) \) has a convergent subsequence, and \( (y_n) \) has a convergent subsequence.
• D. NO subsequence of \( (x_n) \) is convergent, and NO subsequence of \( (y_n) \) is convergent.

Hint:

For oscillating sequences, check for convergent subsequences by analyzing their limiting behavior. A sequence may not converge, but subsequences might.

Answer 1

The Correct Option is B

Step 1: Analyze \( (x_n) \).
The sequence \( x_n = (-1)^n \frac{3n}{n^3} \) simplifies to \( x_n = (-1)^n \frac{3}{n^2} \). As \( n \to \infty \), \( x_n \to 0 \), but since the sequence oscillates due to the factor \( (-1)^n \), it does not converge. Therefore, \( (x_n) \) does not have any convergent subsequences. 
Step 2: Analyze \( (y_n) \).
The sequence \( y_n = \left(4n^3 + (-1)^n 3n^3 \right)^{1/n} \) behaves as: \[ y_n = \left( n^3 \left( 4 + (-1)^n 3 \right) \right)^{1/n}. \] As \( n \to \infty \), the sequence oscillates because of the term \( (-1)^n \), but the overall magnitude tends to 1. Thus, \( y_n \) has a convergent subsequence. Final Answer: \[ \boxed{\text{NO subsequence of } (x_n) \text{ is convergent, and } (y_n) \text{ has a convergent subsequence.}} \]