To analyze the given sequence $a_n$ based on monotonicity, boundedness, and convergence, let's break the problem into its components. The sequence is defined as:

$a_n = \frac{42n^2 + 5n + 3}{6n^2} \cdot \cos(2\pi n + \pi), \quad n \in \mathbb{N}.$

Step 1: Simplify $a_n$

The expression can be simplified: $a_n = \left(\frac{42n^2 + 5n + 3}{6n^2}\right) \cdot \cos(2\pi n + \pi).$

1. Fraction simplification: Divide each term in the numerator by $6n^2$: $\frac{42n^2}{6n^2} + \frac{5n}{6n^2} + \frac{3}{6n^2} = 7 + \frac{5}{6n} + \frac{1}{2n^2}.$ Therefore: $a_n = \left(7 + \frac{5}{6n} + \frac{1}{2n^2}\right) \cdot \cos(2\pi n + \pi).$

2. Cosine term: Since $\cos(2\pi n + \pi) = -1$ for all $n \in \mathbb{N}$, we have: $a_n = -\left(7 + \frac{5}{6n} + \frac{1}{2n^2}\right).$

Step 2: Analyze Monotonicity

The monotonicity depends on how $a_n$ behaves as $n$ increases. For large $n$, the terms $\frac{5}{6n}$ and $\frac{1}{2n^2}$ decrease, meaning $a_n$ approaches $-7$ from above.

However, the exact monotonicity would need the derivative or a difference calculation: $a_{n+1} - a_n = \text{expression involving } n.$ Given the alternating nature of $\cos(2\pi n + \pi)$, the sequence is not monotonic because it oscillates.

Step 3: Check Boundedness

The sequence $a_n$ has a cosine term bounded between $-1$ and $1$, so $a_n$ is also bounded. Specifically: $-\left(7 + \frac{5}{6n} + \frac{1}{2n^2}\right) \leq a_n \leq 0.$ Thus, $a_n$ is bounded above by $0$ and bounded below by $-7$ (as $n \to \infty$).

Step 4: Determine Convergence

For large $n$, the terms $\frac{5}{6n}$ and $\frac{1}{2n^2}$ vanish, leaving: $a_n \to -7.$ Thus, the sequence converges to $-7$.

Final Answers:

1. Monotonicity: The sequence is not monotonic.
2. Boundedness: The sequence is bounded.
3. Convergence: The sequence converges to $-7$: $\lim_{n \to \infty} a_n = -7.$

Would you like further clarification or details? Here are some questions to explore further:

1. How is the behavior of the cosine function crucial in determining boundedness?
2. Why does the oscillatory nature prevent monotonicity?
3. What steps would you take to rigorously calculate $a_{n+1} - a_n$?
4. Can you confirm boundedness for a similar sequence with a sine function?
5. How does the leading term $7$ dominate for large $n$?

Tip: When dealing with sequences involving trigonometric terms, always analyze their periodic properties alongside the dominant growth terms.