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We are given the sequence \( s_n = x^{1/n} \), where \( 0 < x < 1 \), and the task is to prove that this sequence converges and find its limit. We're also instructed to use the Monotone Convergence Theorem and consider the subsequence \( s_{2n} \).

### Step 1: Intuition about the limit
The sequence is \( s_n = x^{1/n} \), where \( x \) is a number between 0 and 1. As \( n \to \infty \), the exponent \( \frac{1}{n} \to 0 \), which implies:

\[
s_n = x^{1/n} \to x^0 = 1 \quad \text{as} \quad n \to \infty.
\]

Thus, intuitively, the sequence \( s_n \) seems to converge to 1.

### Step 2: Monotonicity of the sequence
Let's analyze whether the sequence is monotonic. To do this, we can check if the sequence is decreasing or increasing by comparing \( s_n \) and \( s_{n+1} \).

\[
s_{n+1} = x^{1/(n+1)}.
\]

Now, we want to check if \( s_{n+1} \leq s_n \), i.e., if \( x^{1/(n+1)} \leq x^{1/n} \). This is equivalent to:

\[
\left( x^{1/n} \right)^n \leq \left( x^{1/(n+1)} \right)^{n+1}.
\]

Since \( 0 < x < 1 \), the function \( x^{1/n} \) is increasing as \( n \to \infty \), which implies that the sequence is **increasing** and bounded above by 1. By the Monotone Convergence Theorem, the sequence converges.

### Step 3: Subsequence analysis
The hint also suggests considering the subsequence \( s_{2n} = x^{1/(2n)} \). Similar to the main sequence, as \( n \to \infty \), \( \frac{1}{2n} \to 0 \), so \( s_{2n} \to x^0 = 1 \).

### Step 4: Conclusion
The sequence \( s_n = x^{1/n} \) is increasing and bounded above by 1, which means it converges by the Monotone Convergence Theorem. Therefore, the limit of the sequence is:

\[
\lim_{n \to \infty} s_n = 1.
\]

### Would you like more details on any part of the solution?

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Here are 5 follow-up questions to explore further:

1. How does the Monotone Convergence Theorem guarantee convergence in sequences?
2. What happens to the sequence if \( x > 1 \) instead of \( 0 < x < 1 \)?
3. How would the behavior of the sequence change if \( x = 1 \)?
4. Could you extend this analysis to more complex sequences, like \( x^{1/n^k} \) for some integer \( k \)?
5. How could we apply similar reasoning to sequences involving logarithms or trigonometric functions?

**Tip:** Monotonicity is a powerful property that simplifies proving convergence when you can establish a sequence is bounded and either increasing or decreasing.

Fix a number 0 < x < 1 and consider the sequence of real numbers given by s_n = x^{1/n}. Prove the sequence converges and find its limit.

Learn how to prove the convergence of the sequence s_n = x^{1/n} for 0 < x < 1 using the Monotone Convergence Theorem. This detailed solution explores the subsequence s_{2n} and the sequence's behavior as n approaches infinity. Suitable for undergraduate students in calculus or real analysis.

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