Question

$$\text{Let  } a_n = \begin{cases} 2 + \dfrac{(-1)^{\frac{n-1}{2}}}{n}, & \text{if } n \text{ is odd} \\ 1 + \dfrac{1}{2^n}, & \text{if } n \text{ is even} \end{cases}, \quad n \in \mathbb{N}.$$
Then which one of the following is TRUE?

• A. \( \sup \{a_n \mid n \in \mathbb{N} \} = 3 \) and \( \inf \{a_n \mid n \in \mathbb{N} \} = 1 \)
• B. \( \lim \inf (a_n) = \lim \sup (a_n) = \frac{3}{2} \)
• C. \( \sup \{a_n \mid n \in \mathbb{N} \} = 2 \) and \( \inf \{a_n \mid n \in \mathbb{N} \} = 1 \)
• D. \( \lim \inf (a_n) = 1 \) and \( \lim \sup (a_n) = 3 \)

Hint:

For sequences with alternating behavior, the \( \lim \inf \) and \( \lim \sup \) represent the smallest and largest accumulation points of the sequence.

Answer 1

The Correct Option is A

Step 1: Analyze the sequence for odd values of n

For odd $n$, we have $n = 2k+1$ where $k \geq 0$, so $\frac{n-1}{2} = k$.

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

• When $k$ is even (i.e., $n = 1, 5, 9, 13, ...$): $a_n = 2 + \frac{1}{n}$
• When $k$ is odd (i.e., $n = 3, 7, 11, 15, ...$): $a_n = 2 - \frac{1}{n}$

Let's compute some values:

• $a_1 = 2 + \frac{1}{1} = 3$
• $a_3 = 2 - \frac{1}{3} = \frac{5}{3} \approx 1.667$
• $a_5 = 2 + \frac{1}{5} = \frac{11}{5} = 2.2$
• $a_7 = 2 - \frac{1}{7} = \frac{13}{7} \approx 1.857$

Step 2: Analyze the sequence for even values of n

For even $n$: $$a_n = 1 + \frac{1}{2^n}$$

• $a_2 = 1 + \frac{1}{4} = 1.25$
• $a_4 = 1 + \frac{1}{16} = 1.0625$
• $a_6 = 1 + \frac{1}{64} \approx 1.0156$
• $a_8 = 1 + \frac{1}{256} \approx 1.0039$

Step 3: Find supremum and infimum

From our analysis:

• The maximum value occurs at $n=1$: $a_1 = 3$
• As $n \to \infty$ (even): $a_n \to 1^+$
• As $n \to \infty$ (odd): $a_n \to 2$

For odd $n$:

• Maximum: $a_1 = 3$
• Values approach 2 from both sides but stay in $(2-1/n, 2+1/n)$

For even $n$:

• Values are all greater than 1 and approach 1 from above
• Minimum among even terms approaches 1 (but never equals 1)

Therefore: $$\sup{a_n \mid n \in \mathbb{N}} = 3$$ $$\inf{a_n \mid n \in \mathbb{N}} = 1$$

(The infimum is 1 because the even subsequence approaches 1, though it never reaches it)

Step 4: Find lim sup and lim inf

$$\limsup_{n \to \infty} a_n = \max{\left\{\lim_{n \to \infty, n \text{ odd}} a_n, \lim_{n \to \infty, n \text{ even}} a_n\right\}} = \max{2, 1} = 2$$

$$\liminf_{n \to \infty} a_n = \min{\left\{\lim_{n \to \infty, n \text{ odd}} a_n, \lim_{n \to \infty, n \text{ even}} a_n\right\}} = \min{2, 1} = 1$$

Answer: Option (A)