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The correct answer is:

**(a) \(\langle a_n \rangle\) has always a convergent subsequence.**

This is known as the **Bolzano-Weierstrass theorem**, which states that every bounded sequence of real numbers has a convergent subsequence. It doesn't require the sequence to be bounded in the options given here.

Do you want more details or have any questions?

Here are 5 related questions to expand your understanding:

1. What is the Bolzano-Weierstrass theorem, and why is it important?
2. Can an unbounded sequence have a convergent subsequence?
3. How does the Cauchy criterion relate to the convergence of sequences?
4. What are some examples of sequences that do not converge but have convergent subsequences?
5. Can you explain the difference between a sequence being bounded and being convergent?

**Tip:** For a sequence to converge, every subsequence must converge to the same limit.

Let ⟨a_n⟩ be a sequence of real numbers, then:
a) ⟨a_n⟩ has always a convergent subsequence.
b) ⟨a_n⟩ is convergent if its subsequences ⟨a_2n⟩ and ⟨a_2n+1⟩ are convergent.
c) ⟨a_n⟩ is convergent if its subsequences ⟨a_2n⟩, ⟨a_2n+1⟩, and ⟨a_3n⟩ are convergent.
d) ⟨a_n⟩ is convergent if it is bounded.

Understand the properties of sequences of real numbers, focusing on subsequences and convergence. Explore how the Bolzano-Weierstrass theorem applies and when a sequence has a convergent subsequence. Suitable for first-year undergraduate students studying real analysis.

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