We are tasked with determining whether the sequence converges or diverges, and if it converges, finding its limit. The given sequence is:

$a_n = \frac{3n^2 + 1}{\sqrt{9n^3 + 4n}}$

Step-by-step analysis:

1. Simplify the sequence for large $n$: We are interested in the behavior of the sequence as $n \to \infty$. For large $n$, the highest degree terms will dominate the behavior of the sequence, so we focus on the leading terms in the numerator and denominator.

   • In the numerator, $3n^2 + 1$, the term $3n^2$ dominates as $n \to \infty$.
   • In the denominator, $\sqrt{9n^3 + 4n}$, the term $\sqrt{9n^3} = 3n^{3/2}$ dominates as $n \to \infty$.

   So, we approximate the sequence by the leading terms: $a_n \approx \frac{3n^2}{3n^{3/2}} = n^{1/2} = \sqrt{n}$

2. Limit of the sequence: Since $a_n \approx \sqrt{n}$ as $n \to \infty$, the sequence behaves like $\sqrt{n}$, which grows without bound. Therefore, the sequence diverges to infinity.

Conclusion:

The sequence $a_n$ diverges as $n \to \infty$.

Would you like any further details or clarification?

Here are five related questions to deepen your understanding:

1. What is the difference between a sequence diverging to infinity and oscillating divergence?
2. How can we determine the rate of growth of sequences for large $n$?
3. What happens if the numerator and denominator had different dominating terms in higher powers of $n$?
4. How would the analysis change if the denominator's degree were less than that of the numerator?
5. Can you find an example of a sequence where both the numerator and denominator have the same degree in $n$?

Tip: When analyzing sequences, always start by focusing on the highest degree terms, as they often determine the behavior at infinity.