The sequence given is $a_n = (-1)^n$.

Let's analyze whether the sequence converges:

1. Behavior of the sequence:

The term $(-1)^n$ alternates between $1$ and $-1$ for successive values of $n$. Specifically:

• If $n$ is even, $(-1)^n = 1$.
• If $n$ is odd, $(-1)^n = -1$.

Thus, the sequence alternates as follows:

• For $n = 1$, $a_1 = -1$
• For $n = 2$, $a_2 = 1$
• For $n = 3$, $a_3 = -1$
• For $n = 4$, $a_4 = 1$, and so on.

2. Convergence:

A sequence converges if, as $n$ approaches infinity, the terms of the sequence approach a single fixed value (the limit).

In this case, the sequence $a_n = (-1)^n$ does not approach a single value as $n \to \infty$. Instead, it continues to alternate between $1$ and $-1$, never settling on one particular value.

Therefore, the sequence does not converge.

Conclusion:

The sequence $a_n = (-1)^n$ does not converge, and hence, it has no limit.

Would you like further details or have any questions?

Here are five related questions you might consider:

1. What is the definition of a convergent sequence?
2. Can a sequence that alternates between two values ever converge?
3. How do we mathematically prove that a sequence converges?
4. What is an example of a sequence that converges to a limit?
5. What is the difference between oscillating and converging sequences?

Tip: For a sequence to converge, the values of the sequence must get arbitrarily close to a specific number as $n \to \infty$.