Question

The least upper bound (lub) and greatest lower bound (glb) of the set \( \{1 + ((-1)^n / n) ; n\in \mathbb{N\} \) are:}

• A. lub = 1, glb = -1
• B. lub = 3/2, glb = 1
• C. lub = 3/2, glb = 0
• D. lub = 1, glb = 0

Hint:

Exam Tip:
When finding lub and glb of alternating sequences:

• Separate into even and odd subsequences.
• Find the maximum and minimum of the entire sequence.
• The limit points (like 1 in this case) are not the lub or glb if the sequence takes values beyond them.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The set is defined by \(a_n = 1 + \frac{(-1)^n}{n}\) for \(n \in \mathbb{N}\) (natural numbers, typically \(n = 1, 2, 3, \dots\)).
We need to find the least upper bound (supremum) and greatest lower bound (infimum) of this set.

Step 2: Analyzing the Sequence:
Let's list the first few terms to see the pattern: \[ \begin{aligned} n = 1: &\quad 1 + \frac{(-1)^1}{1} = 1 - 1 = 0 n = 2: &\quad 1 + \frac{(-1)^2}{2} = 1 + \frac{1}{2} = \frac{3}{2} n = 3: &\quad 1 + \frac{(-1)^3}{3} = 1 - \frac{1}{3} = \frac{2}{3} n = 4: &\quad 1 + \frac{(-1)^4}{4} = 1 + \frac{1}{4} = \frac{5}{4} n = 5: &\quad 1 + \frac{(-1)^5}{5} = 1 - \frac{1}{5} = \frac{4}{5} \end{aligned} \] The sequence alternates between two subsequences:
• For even \(n\): \(a_{2k} = 1 + \frac{1}{2k}\), which decreases and approaches 1 from above as \(k \to \infty\).
• For odd \(n\): \(a_{2k-1} = 1 - \frac{1}{2k-1}\), which increases and approaches 1 from below as \(k \to \infty\).

Step 3: Finding the lub and glb:
The maximum value in the set occurs at \(n = 2\), where \(a_2 = \frac{3}{2}\).
Thus, \(\text{lub} = \frac{3}{2}\).
The minimum value in the set occurs at \(n = 1\), where \(a_1 = 0\).
Thus, \(\text{glb} = 0\).
The sequence approaches 1 but never reaches it for \(n > 2\).
Therefore, the set is bounded and its extremes are exactly 0 and 3/2.

Step 4: Final Answer:
lub = 3/2 and glb = 0. Therefore, option (C) is correct.