To find the convergence of the sequence given by

$a_n = \frac{n + (-1)^n}{2n + (-1)^n},$

we'll explore its behavior as $n$ approaches infinity.

Step 1: Analyze the structure

The sequence involves the term $(-1)^n$, which alternates between $+1$ and $-1$ depending on whether $n$ is even or odd.

Thus, we need to consider two cases:

1. When $n$ is even, $(-1)^n = 1$
2. When $n$ is odd, $(-1)^n = -1$

Case 1: $n$ is even

If $n$ is even, then $(-1)^n = 1$, and the expression becomes:

$a_n = \frac{n + 1}{2n + 1}.$

As $n \to \infty$, both the numerator and denominator grow large, but the ratio tends toward:

$\lim_{n \to \infty} \frac{n + 1}{2n + 1} = \lim_{n \to \infty} \frac{n(1 + \frac{1}{n})}{n(2 + \frac{1}{n})} = \frac{1}{2}.$

Case 2: $n$ is odd

If $n$ is odd, then $(-1)^n = -1$, and the expression becomes:

$a_n = \frac{n - 1}{2n - 1}.$

Similarly, as $n \to \infty$, we compute the limit:

$\lim_{n \to \infty} \frac{n - 1}{2n - 1} = \lim_{n \to \infty} \frac{n(1 - \frac{1}{n})}{n(2 - \frac{1}{n})} = \frac{1}{2}.$

Step 2: Conclusion

In both cases (whether $n$ is even or odd), the sequence approaches the same limit:

$\lim_{n \to \infty} a_n = \frac{1}{2}.$

Thus, the sequence converges to $\frac{1}{2}$ as $n \to \infty$.

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Would you like more details or explanations on any of these steps? Here are 5 related questions you might find interesting:

1. How do you determine if a sequence converges in general?
2. What happens if a sequence has different limits for even and odd $n$?
3. Can we use L'Hôpital's rule for solving limits of sequences?
4. What are some common techniques for analyzing the convergence of alternating sequences?
5. How can we apply this approach to other similar sequences?

Tip: Always break down complex expressions into simpler cases, like analyzing even and odd behavior, to gain insights into their limits.