We are tasked with proving that the sequence $a_n = 1 - \frac{(-1)^n}{n}$ converges.

Step 1: Understanding the sequence

The sequence is defined as: $a_n = 1 - \frac{(-1)^n}{n}.$ Here, $(-1)^n$ alternates between $1$ and $-1$ depending on whether $n$ is even or odd. Therefore, we can break this into two subsequences based on whether $n$ is even or odd:

• When $n$ is even, $(-1)^n = 1$, so the sequence becomes: $a_n = 1 - \frac{1}{n}.$
• When $n$ is odd, $(-1)^n = -1$, so the sequence becomes: $a_n = 1 - \frac{-1}{n} = 1 + \frac{1}{n}.$

Step 2: Analyze the behavior as $n \to \infty$

Now let's analyze the behavior of these subsequences as $n$ becomes very large.

• When $n$ is even: $a_n = 1 - \frac{1}{n} \to 1$ as $n \to \infty$.
• When $n$ is odd: $a_n = 1 + \frac{1}{n} \to 1$ as $n \to \infty$.

In both cases, the value of $a_n$ approaches 1 as $n \to \infty$, regardless of whether $n$ is even or odd.

Step 3: Conclusion

Since both subsequences converge to the same limit, we conclude that the sequence $a_n = 1 - \frac{(-1)^n}{n}$ converges to 1.

Thus, the sequence converges, and: $\lim_{n \to \infty} \left( 1 - \frac{(-1)^n}{n} \right) = 1.$

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Would you like more details on any step or have any further questions?

Here are five related questions:

1. How do we define the limit of a sequence?
2. Can a sequence converge to different limits based on subsequences?
3. What is the definition of a convergent sequence?
4. How do alternating sequences behave in terms of convergence?
5. How would the sequence change if the term $(-1)^n$ were replaced with another function?

Tip: Breaking a sequence into subsequences can simplify the analysis of convergence, especially for alternating or periodic terms.