is the sequence cos(npi/2) divergent or convergent

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.

Is the sequence cos(npi/2) divergent or convergent?

Learn why the sequence cos(nπ/2) is divergent. This problem demonstrates how sequences that oscillate between fixed values without approaching a single limit are classified as divergent. Suitable for high school students studying sequences and trigonometric functions.

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