To determine whether the sequence $a_n = \ln(3n + 1) - \ln(6n)$ converges or diverges, we need to find the limit:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(\ln(3n + 1) - \ln(6n)\right).$

Step 1: Simplify the expression

Using the properties of logarithms, we can combine the terms:

$a_n = \ln\left(\frac{3n + 1}{6n}\right).$

Step 2: Simplify the fraction inside the logarithm

$\frac{3n + 1}{6n} = \frac{3n}{6n} + \frac{1}{6n} = \frac{1}{2} + \frac{1}{6n}.$

As $n \to \infty$, the term $\frac{1}{6n} \to 0$. Therefore:

$\lim_{n \to \infty} \frac{3n + 1}{6n} = \frac{1}{2}.$

Step 3: Take the limit

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \ln\left(\frac{3n + 1}{6n}\right) = \ln\left(\frac{1}{2}\right) = -\ln(2).$

Conclusion:

The limit exists and is equal to $-\ln(2)$. Therefore, the sequence converges to $-\ln(2)$.

Answer: $\lim_{n \to \infty} a_n = -\ln(2),$ and the sequence converges.

Would you like more details or have any questions?

Related Questions:

1. What is the general approach to finding the limit of a sequence involving logarithms?
2. How can you use L'Hopital's Rule to find the limit of a sequence if needed?
3. What does it mean for a sequence to converge, and how is it different from diverging?
4. Can you explain how properties of logarithms help simplify expressions?
5. How do we handle sequences where the terms inside a logarithm approach zero?

Tip: When working with logarithmic sequences, always try to simplify expressions using logarithmic properties to make the limit evaluation easier.