Let's analyze the sequence $a_n = \cos\left(\pi^{-n}\right)$:

1. Bounded Above and Below:

   • The function $\cos(x)$ always takes values between $-1$ and $1$ for any real number $x$.
   • Therefore, for all $n$, $a_n = \cos(\pi^{-n})$ must also be between $-1$ and $1$.
   • So the sequence is bounded above (by 1) and bounded below (by -1).
   • Therefore, the sequence is bounded.

2. Monotonicity:

   • Consider the argument inside the cosine, $\pi^{-n}$, which decreases as $n$ increases.
   • Since $\cos(x)$ is a periodic function, it's not guaranteed that the values of $\cos(\pi^{-n})$ will decrease or increase consistently as $n$ increases. This function oscillates, meaning the sequence is neither non-increasing nor non-decreasing.
   • Thus, the sequence is not monotone.

3. Convergence or Divergence:

   • As $n \to \infty$, $\pi^{-n} \to 0$.
   • Since $\cos(0) = 1$, the sequence converges to 1.
   • Hence, the sequence converges.

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Correct True Statements:

• The sequence is bounded above.
• The sequence is bounded below.
• The sequence is bounded.
• The sequence converges.

False Statements:

• The sequence is non-increasing.
• The sequence is non-decreasing.
• The sequence is monotone.
• The sequence diverges.

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Do you have any questions or would you like further clarification on any part?

Here are 5 related questions to consider:

1. What does it mean for a sequence to be bounded?
2. How do we formally prove a sequence is convergent?
3. What is the difference between a monotone and oscillating sequence?
4. Can a sequence be bounded and divergent at the same time?
5. How does the periodic nature of cosine affect the behavior of sequences?

Tip: When studying sequences, always check if the sequence has a limit as $n \to \infty$, as this helps in determining whether the sequence converges or diverges.