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### Problem 8:
*If \((s_n)\) is a subsequence of \((t_n)\) and \((t_n)\) is a subsequence of \((s_n)\), can we conclude that \((s_n) = (t_n)\)? Prove or give a counterexample.*

#### Solution:
We need to decide if two sequences that are subsequences of each other must be identical. A counterexample would disprove this claim.

**Counterexample:**
Consider the sequence \((s_n) = (-1)^n\), which alternates between \(1\) and \(-1\). Now, define:
- \((t_n)\) as the subsequence of \((s_n)\) where \(t_n = s_{2n} = 1\) for all \(n\), i.e., it takes only the positive terms from \((s_n)\).
- \((s_n)\) as the subsequence of \((t_n)\), where \(s_n = t_{2n} = 1\), i.e., it also only includes positive terms.

Even though both are subsequences of each other, they are not identical because one sequence consists of both \(1\) and \(-1\), while the other only consists of \(1\).

Thus, **no**, we cannot conclude that \((s_n) = (t_n)\) if they are subsequences of each other.

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### Problem 9:
*Let \((s_n)\) be a bounded sequence and suppose that \(\liminf s_n = \limsup s_n = s\). Prove that \((s_n)\) is convergent and that \(\lim s_n = s\).*

#### Solution:
We are given that the sequence \((s_n)\) is bounded, and \(\liminf s_n = \limsup s_n = s\). We want to prove that the sequence converges to \(s\).

1. **Boundedness**: Since \((s_n)\) is bounded, it means there exists some real numbers \(m\) and \(M\) such that \(m \leq s_n \leq M\) for all \(n\).
   
2. **Equality of \(\liminf\) and \(\limsup\)**: The limit inferior (\(\liminf s_n\)) is the largest value such that any subsequence converging below it must converge to a limit smaller than or equal to this value. Similarly, the limit superior (\(\limsup s_n\)) is the smallest value such that any subsequence converging above it must converge to a limit larger than or equal to it.

   Since \(\liminf s_n = \limsup s_n = s\), this implies that all subsequential limits must converge to \(s\). No subsequence can converge to anything other than \(s\), so the entire sequence must converge to \(s\).

Thus, \((s_n)\) converges to \(s\).

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### Problem 10:
*Suppose that \(x > 1\). Prove that \(\lim_{n \to \infty} x^{1/n} = 1\).*

#### Solution:
We need to prove that as \(n \to \infty\), \(x^{1/n}\) approaches 1 for any \(x > 1\).

1. Rewrite \(x^{1/n}\) as \(\exp\left(\frac{\ln(x)}{n}\right)\).
   
2. As \(n \to \infty\), the term \(\frac{\ln(x)}{n}\) approaches 0 because \(\ln(x)\) is a constant and \(n\) grows indefinitely.

3. Therefore, \(\exp\left(\frac{\ln(x)}{n}\right)\) approaches \(\exp(0) = 1\).

Thus, we have \(\lim_{n \to \infty} x^{1/n} = 1\).

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Do you need further details or clarifications on any of these problems?

### Follow-up Questions:
1. What is the formal definition of a subsequence?
2. Can you give another example where two subsequences are not equal?
3. How does the definition of \(\liminf\) and \(\limsup\) help in proving convergence?
4. What is the intuition behind the logarithmic approach in Problem 10?
5. How does boundedness of a sequence guarantee the existence of subsequential limits?

### Tip:
Always check if a counterexample exists when a statement seems overly general, as it can save time compared to attempting a direct proof.

8. If (s_n) is a subsequence of (t_n) and (t_n) is a subsequence of (s_n), can we conclude that (s_n) = (t_n)? Prove or give a counterexample. 9. Let (s_n) be a bounded sequence and suppose that lim inf s_n = lim sup s_n = s. Prove that (s_n) is convergent and that lim s_n = s. 10. Suppose that x > 1. Prove that lim x^{1/n} = 1.

Explore mathematical problems involving subsequences and convergence. This includes proving the relationship between subsequences, the limit of a bounded sequence with equal lim inf and lim sup, and the limit of x^(1/n) as n approaches infinity. Suitable for undergraduate mathematics students.

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