Question:
Which of the following is TRUE?
Which of the following is TRUE?
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A sequence having a convergent subsequence is always bounded, but not necessarily Cauchy.
Updated On: Nov 20, 2025
- Every sequence that has a convergent subsequence is a Cauchy sequence
- The sequence \( \{sin n\} \) has a convergent subsequence
- Every sequence that has a convergent subsequence is a bounded sequence
- The sequence \( \{n \cos \frac{1}{n}\} \) has a convergent subsequence
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the problem.
By the Bolzano-Weierstrass Theorem, every sequence in \( \mathbb{R} \) has a convergent subsequence if and only if the sequence is bounded. Therefore, if a sequence has a convergent subsequence, it must be bounded.
Step 2: Analyzing the options.
(A) Every sequence that has a convergent subsequence is a Cauchy sequence: This is not true. A sequence can have a convergent subsequence without being Cauchy.
(B) Every sequence that has a convergent subsequence is a bounded sequence: This is true, as mentioned by the Bolzano-Weierstrass Theorem.
(C) The sequence \( \{sin n\} \) has a convergent subsequence: This is true. \( \{sin n\} \) is bounded and has a convergent subsequence.
(D) The sequence \( \{n \cos \frac{1}{n}\} \) has a convergent subsequence: This is true, but it is not the most general answer.
Step 3: Conclusion.
Thus, the correct answer is (B).
By the Bolzano-Weierstrass Theorem, every sequence in \( \mathbb{R} \) has a convergent subsequence if and only if the sequence is bounded. Therefore, if a sequence has a convergent subsequence, it must be bounded.
Step 2: Analyzing the options.
(A) Every sequence that has a convergent subsequence is a Cauchy sequence: This is not true. A sequence can have a convergent subsequence without being Cauchy.
(B) Every sequence that has a convergent subsequence is a bounded sequence: This is true, as mentioned by the Bolzano-Weierstrass Theorem.
(C) The sequence \( \{sin n\} \) has a convergent subsequence: This is true. \( \{sin n\} \) is bounded and has a convergent subsequence.
(D) The sequence \( \{n \cos \frac{1}{n}\} \) has a convergent subsequence: This is true, but it is not the most general answer.
Step 3: Conclusion.
Thus, the correct answer is (B).
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