The given sequence is:

$a_n = \frac{6n}{\sqrt{49n^3 + 6}}$

We need to determine if the sequence converges or diverges, and if it converges, find its limit.

Step 1: Simplify the expression for large $n$

As $n \to \infty$, the dominant term in the denominator is $49n^3$, so we can approximate the expression:

$a_n = \frac{6n}{\sqrt{49n^3 + 6}} \approx \frac{6n}{\sqrt{49n^3}} = \frac{6n}{\sqrt{49}n^{3/2}} = \frac{6n}{7n^{3/2}}$

Step 2: Simplify the leading terms

The simplified expression is:

$a_n \approx \frac{6}{7n^{1/2}} = \frac{6}{7\sqrt{n}}$

Step 3: Find the limit as $n \to \infty$

As $n$ approaches infinity, $\frac{6}{7\sqrt{n}}$ approaches 0 because the denominator grows without bound. Therefore:

$\lim_{n \to \infty} a_n = 0$

Conclusion:

The sequence converges, and its limit is 0.

So:

• $\lim_{n \to \infty} a_n = 0$
• The sequence converges.

Would you like more details on any step? Here are five related questions you might find interesting:

1. What happens if the highest degree of the numerator and denominator are equal?
2. How does the behavior of sequences change if the degree of the numerator is larger than that of the denominator?
3. Can we use L'Hopital's Rule to find limits of sequences like this?
4. What are the common methods for determining sequence divergence?
5. How can you check for convergence using the ratio test for sequences?

Tip: When analyzing limits of sequences with radicals, always check the dominant term for large values of $n$, as it often simplifies the calculation.