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To determine the convergence or divergence of the series 

\[
\sum_{n=1}^{\infty} \frac{\sqrt[6]{n}}{3n^2 + 6},
\]

we can analyze the behavior of the terms as \( n \to \infty \) and determine if the series converges by applying comparison tests or the Limit Comparison Test. Here’s a step-by-step approach:

1. **Rewrite the term**: The term can be rewritten as:
   \[
   \frac{\sqrt[6]{n}}{3n^2 + 6} = \frac{n^{1/6}}{3n^2 + 6}.
   \]

2. **Identify dominant behavior**: As \( n \to \infty \), the dominant term in the denominator is \( 3n^2 \), so we approximate the term as:
   \[
   \frac{n^{1/6}}{3n^2} = \frac{1}{3} \cdot \frac{1}{n^{2 - 1/6}} = \frac{1}{3} \cdot \frac{1}{n^{11/6}}.
   \]

3. **Compare with a \( p \)-series**: This expression behaves similarly to a \( p \)-series, \( \sum \frac{1}{n^{p}} \), with \( p = \frac{11}{6} \). For a \( p \)-series, the series converges if \( p > 1 \). Since \( \frac{11}{6} > 1 \), the series \( \sum \frac{1}{n^{11/6}} \) converges.

4. **Limit Comparison Test**: To formally confirm, we use the Limit Comparison Test with \( b_n = \frac{1}{n^{11/6}} \). Compute:
   \[
   \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n^{1/6}}{3n^2 + 6}}{\frac{1}{n^{11/6}}} = \lim_{n \to \infty} \frac{n^{1/6} \cdot n^{11/6}}{3n^2 + 6} = \lim_{n \to \infty} \frac{n^2}{3n^2 + 6}.
   \]
   Simplifying this gives:
   \[
   \lim_{n \to \infty} \frac{n^2}{3n^2 + 6} = \frac{1}{3}.
   \]
   Since this limit is a positive finite constant, the series \( \sum a_n \) converges if \( \sum b_n \) converges. Since \( \sum \frac{1}{n^{11/6}} \) converges, \( \sum_{n=1}^{\infty} \frac{\sqrt[6]{n}}{3n^2 + 6} \) also converges.

### Conclusion:
The series \( \sum_{n=1}^{\infty} \frac{\sqrt[6]{n}}{3n^2 + 6} \) **converges**.

Would you like further details or have any questions?

### Related Questions
1. How does the Limit Comparison Test work in determining convergence?
2. What are \( p \)-series, and why do they converge when \( p > 1 \)?
3. Could this series converge using a different test?
4. What happens to the convergence of this series if the exponent of \( n \) in the denominator changes?
5. How does the behavior of \( n^{1/6} \) affect the convergence of the series?

### Tip
For series with dominant polynomial terms in the denominator, the Limit Comparison Test with a \( p \)-series often simplifies convergence analysis.

Determine convergence or divergence of \( \sum_{n=1}^{\infty} \frac{\sqrt[6]{n}}{3n^2 + 6} \).

Explore the convergence or divergence of the series \( \sum \frac{\sqrt[6]{n}}{3n^2 + 6} \) using the Limit Comparison Test and p-series analysis. This problem involves concepts in calculus, including comparison with a p-series, for students in advanced mathematics courses.

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