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We are given the sequence \( a_n = \sqrt[n]{n} \), and we are tasked with computing the first six terms, rounding the results to four decimal places.

### The first six terms:
1. \( a_1 = \sqrt[1]{1} = 1.0000 \)
2. \( a_2 = \sqrt[2]{2} = \sqrt{2} \approx 1.4142 \)
3. \( a_3 = \sqrt[3]{3} \approx 1.4422 \)
4. \( a_4 = \sqrt[4]{4} = \sqrt[2]{2} \approx 1.1892 \)
5. \( a_5 = \sqrt[5]{5} \approx 1.1487 \)
6. \( a_6 = \sqrt[6]{6} \approx 1.1225 \)

### Convergence of the Sequence:
We are asked whether the sequence converges or diverges. As \( n \to \infty \), \( \sqrt[n]{n} \) approaches 1 because the root grows faster than the value of \( n \), which causes the expression to get closer to 1. Hence, the sequence converges to 1.

So, the **limit** is \( \boxed{1} \).

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Would you like more details or have any questions?

### Relative Questions:
1. What is the general behavior of \( \sqrt[n]{n} \) as \( n \) increases?
2. How can we prove that \( \lim_{n \to \infty} \sqrt[n]{n} = 1 \) rigorously?
3. How does the sequence \( \sqrt[n]{n} \) compare to other common sequences, like \( \frac{1}{n} \)?
4. Can you find another sequence where the root increases faster than the base, leading to divergence?
5. What happens to the sequence if \( n \) grows extremely large, say in the thousands?

### Tip:
Sequences involving roots tend to converge slowly. Understanding how different parts of the expression behave (like the base and the root) helps in analyzing limits more intuitively.

Compute the first six terms of the sequence {a_n} = nth root of n. (Round your answers to four decimal places.) If the sequence converges, find its limit.

Learn how to compute the first six terms of the sequence {a_n} = nth root of n, rounded to four decimal places. Understand the convergence of the sequence and find its limit. Suitable for students in Grades 10-12.

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