The sequence in question is $a_n = (-1)^n$. Let's analyze its convergence.

Step-by-Step Analysis:

1. General Term: The sequence is defined as: $a_n = (-1)^n$ Here, $(-1)^n$ alternates between $-1$ and $1$ depending on whether $n$ is odd or even, respectively.

2. Behavior of the Sequence:

   • When $n$ is even: $a_n = 1$
   • When $n$ is odd: $a_n = -1$

   Therefore, the sequence does not settle on a single value as $n$ increases; instead, it continues to alternate between $-1$ and $1$.

3. Convergence:

   • For a sequence to converge, the terms must approach a single value (the limit) as $n$ tends to infinity.
   • However, in this case, the sequence does not approach a single value because it keeps oscillating between $-1$ and $1$.

4. Conclusion: The sequence $a_n = (-1)^n$ does not converge because it does not settle at a single value as $n$ increases.

Summary:

The sequence $a_n = (-1)^n$ does not converge.

Would you like more details or have any questions?

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Related Questions:

1. What are necessary conditions for a sequence to converge?
2. How can you determine if a series with alternating signs converges?
3. What is the difference between pointwise and uniform convergence?
4. How does the limit of a subsequence relate to the convergence of the entire sequence?
5. What are some examples of sequences that oscillate but still converge?
6. How does the concept of divergence apply to sequences?
7. Can a sequence be bounded but not convergent?
8. What is the significance of the Bolzano-Weierstrass theorem in analyzing sequences?

Tip: When dealing with sequences, a key indicator of convergence is whether the sequence values approach a fixed limit. Oscillating sequences like $(-1)^n$ typically do not converge.