1. **Problem 1:** Find $\limsup_{n\to\infty} \frac{2(-1)^n}{n}$ and $\liminf_{n\to\infty} \frac{2(-1)^n}{n}$. 2. **Problem 2:** Find $\limsup_{n\to\infty} \frac{3(n+1)(-1)^n}{n}$ and $\liminf_{n\to\infty} \frac{3(n+1)(-1)^n}{n}$. 3. **Problem 3:** Given the sequence $x_n = \begin{cases} \frac{-4}{n} & \text{if } n \text{ even} \\ \frac{4n-1}{n} & \text{if } n \text{ odd} \end{cases}$, define $a_n = \sup\{x_k : k \ge n\}$ and $b_n = \inf\{x_k : k \ge n\}$. Find $a_n$, $b_n$ for even and odd $n$, then find $\limsup x_n$ and $\liminf x_n$. --- ### Step 1: Analyze $\limsup$ and $\liminf$ of $\frac{2(-1)^n}{n}$ - The sequence is $x_n = \frac{2(-1)^n}{n}$. - As $n \to \infty$, $\frac{1}{n} \to 0$. - The term $(-1)^n$ oscillates between $1$ and $-1$. So the sequence terms alternate between positive and negative values approaching zero: $$x_{2m} = \frac{2(1)}{2m} = \frac{2}{2m} = \frac{1}{m} \to 0^+$$ $$x_{2m+1} = \frac{2(-1)}{2m+1} = -\frac{2}{2m+1} \to 0^-$$ - The $\limsup$ is the limit of the supremum of the tail, which is the limit of the positive subsequence: $$\limsup_{n\to\infty} x_n = 0$$ - The $\liminf$ is the limit of the infimum of the tail, which is the limit of the negative subsequence: $$\liminf_{n\to\infty} x_n = 0$$ ### Step 2: Analyze $\limsup$ and $\liminf$ of $\frac{3(n+1)(-1)^n}{n}$ Rewrite the sequence: $$y_n = \frac{3(n+1)(-1)^n}{n} = 3(-1)^n \left(1 + \frac{1}{n}\right)$$ - As $n \to \infty$, $1 + \frac{1}{n} \to 1$. - The sequence oscillates between approximately $3$ and $-3$: $$y_{2m} \to 3$$ $$y_{2m+1} \to -3$$ - So: $$\limsup_{n\to\infty} y_n = 3$$ $$\liminf_{n\to\infty} y_n = -3$$ ### Step 3: Analyze sequence $x_n$ defined by cases $$x_n = \begin{cases} \frac{-4}{n} & n \text{ even} \\ \frac{4n - 1}{n} & n \text{ odd} \end{cases}$$ - For even $n$, $x_n = -\frac{4}{n} \to 0^-$ as $n \to \infty$. - For odd $n$, $$x_n = \frac{4n - 1}{n} = 4 - \frac{1}{n} \to 4^-$$ #### Define $a_n = \sup\{x_k : k \ge n\}$ and $b_n = \inf\{x_k : k \ge n\}$ - For **even** $n$: - The supremum of the tail is dominated by the odd terms which approach 4 from below, so: $$a_n = 4 - \frac{1}{n+1}$$ (since the next odd number after even $n$ is $n+1$) - The infimum is the smallest even term from $n$ onwards, which is $-\frac{4}{n}$ (since $-4/n$ increases towards 0 from below): $$b_n = -\frac{4}{n}$$ - For **odd** $n$: - The supremum is the current odd term or later odd terms, which approach 4: $$a_n = 4 - \frac{1}{n}$$ - The infimum is the smallest even term from $n$ onwards, which is $-\frac{4}{n+1}$ (next even number after odd $n$ is $n+1$): $$b_n = -\frac{4}{n+1}$$ ### Step 4: Find $\limsup x_n$ and $\liminf x_n$ - By definition: $$\limsup x_n = \lim_{n\to\infty} a_n = 4$$ $$\liminf x_n = \lim_{n\to\infty} b_n = 0$$ --- **Final answers:** $$\limsup_{n\to\infty} \frac{2(-1)^n}{n} = 0$$ $$\liminf_{n\to\infty} \frac{2(-1)^n}{n} = 0$$ $$\limsup_{n\to\infty} \frac{3(n+1)(-1)^n}{n} = 3$$ $$\liminf_{n\to\infty} \frac{3(n+1)(-1)^n}{n} = -3$$ For $x_n$: - For even $n$: $$a_n = 4 - \frac{1}{n+1}$$ $$b_n = -\frac{4}{n}$$ - For odd $n$: $$a_n = 4 - \frac{1}{n}$$ $$b_n = -\frac{4}{n+1}$$ - Limits: $$\limsup x_n = 4$$ $$\liminf x_n = 0$$