The given sequence is:

$a_n = \frac{\cos^2(n)}{3^n}$

We are asked to determine whether the sequence converges or diverges, and if it converges, to find the limit.

Step-by-Step Solution:

1. Behavior of the numerator: The function $\cos^2(n)$ oscillates between 0 and 1 for all values of $n$. Specifically, since $\cos(n)$ is periodic and bounded, $\cos^2(n)$ is also bounded, i.e., $0 \leq \cos^2(n) \leq 1 \quad \text{for all } n.$

2. Behavior of the denominator: The term $3^n$ grows exponentially as $n$ increases, since $3^n$ represents an exponentially increasing function with base 3.

3. Limit of the sequence: As $n \to \infty$, the exponential growth of $3^n$ in the denominator will dominate the oscillating $\cos^2(n)$ in the numerator. This leads to the conclusion that the entire sequence tends to 0, because the denominator grows much faster than the numerator can oscillate.

   Formally, we evaluate the limit: $\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\cos^2(n)}{3^n} = 0.$

Conclusion:

The sequence converges and the limit is:

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

Would you like further details or clarifications on any step?

Here are 5 related questions to expand your understanding:

1. What are some other sequences where the exponential term dominates the numerator?
2. How would the limit change if the denominator were $2^n$ instead of $3^n$?
3. Can a sequence with a trigonometric function in the numerator ever diverge?
4. How does the periodic nature of $\cos(n)$ affect the convergence of sequences in general?
5. What is the general approach for determining the limit of sequences involving exponential growth?

Tip: When dealing with sequences involving exponential terms, check if the exponential growth in the denominator can overpower any bounded or oscillating terms in the numerator to determine convergence.