We are given the set:

$A = \left\{ 1/n + (-1)^n : n \in \mathbb{N} \right\}$

Step 1: Understanding the general term

The general term for the set is:

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

This is a sequence where $n \in \mathbb{N}$ (the set of natural numbers). Let's analyze the behavior of $a_n$ for different values of $n$.

• For even $n$, $(-1)^n = 1$, so the term becomes: $a_n = \frac{1}{n} + 1$

• For odd $n$, $(-1)^n = -1$, so the term becomes: $a_n = \frac{1}{n} - 1$

Step 2: Investigating the behavior of the sequence

• Even $n$: $a_n = \frac{1}{n} + 1$ As $n \to \infty$, $\frac{1}{n} \to 0$, so $a_n \to 1$.

• Odd $n$: $a_n = \frac{1}{n} - 1$ As $n \to \infty$, $\frac{1}{n} \to 0$, so $a_n \to -1$.

Thus, for large $n$, the sequence oscillates between values close to 1 (for even $n$) and values close to -1 (for odd $n$).

Step 3: Finding the supremum and infimum

• Supremum: The largest value of $a_n$ occurs for $n = 2$, where: $a_2 = \frac{1}{2} + 1 = 1.5$ Therefore, the supremum of the set $A$ is: $\sup(A) = 1.5$

• Infimum: The smallest value of $a_n$ occurs for $n = 1$, where: $a_1 = \frac{1}{1} - 1 = 0$ As $n \to \infty$, $a_n$ approaches $-1$ for odd $n$, but never reaches it. Hence, the infimum is: $\inf(A) = -1$

Final Answer:

• Supremum: $\sup(A) = 1.5$
• Infimum: $\inf(A) = -1$

Would you like further details or explanations? Here are some related questions to deepen your understanding:

1. What is the difference between supremum and maximum in sets?
2. How do we prove the limits of sequences involving alternating terms?
3. Can we apply similar analysis to more complex alternating sequences?
4. What is the importance of natural numbers ($\mathbb{N}$) in defining sequences?
5. How do different types of limits affect the behavior of sequences?

Tip: When working with sequences, always consider how terms behave for both small and large values of $n$ to understand their limiting behavior.