The sequence $\cos\left(\frac{n \pi}{2}\right)$ is divergent. Let's analyze why:

The sequence $\cos\left(\frac{n \pi}{2}\right)$ depends on the integer values of $n$. By plugging in consecutive values of $n$, we get:

1. For $n = 0$: $\cos(0) = 1$
2. For $n = 1$: $\cos\left(\frac{\pi}{2}\right) = 0$
3. For $n = 2$: $\cos(\pi) = -1$
4. For $n = 3$: $\cos\left(\frac{3\pi}{2}\right) = 0$
5. For $n = 4$: $\cos(2\pi) = 1$

And this pattern repeats: $1, 0, -1, 0, 1, 0, -1, 0, \dots$.

Since the sequence oscillates among the values $1$, $0$, and $-1$ without approaching a single limit, it does not converge. Therefore, the sequence $\cos\left(\frac{n \pi}{2}\right)$ is divergent.

Would you like more details on the reasoning or about sequence convergence in general?

Here are some related questions that might interest you:

1. What are the criteria for determining if a sequence is convergent or divergent?
2. How does the behavior of trigonometric sequences differ from typical algebraic sequences in terms of convergence?
3. Can a sequence be divergent but still bounded?
4. What are examples of other oscillating sequences, and how are they analyzed for convergence?
5. How would we prove formally that a sequence is bounded?

Tip: For trigonometric sequences, analyzing specific values and identifying patterns can quickly reveal convergence or divergence.