Example 1
Solution Firstly, note that we can rewrite each \(a_n\) as \[a_n = (-1)^n \frac{2}{3}\frac{1}{\frac{1}{3n} + 1}.\] Splitting into odd and even cases, we obtain \[a_n = \begin{cases} \frac{2}{3}\frac{1}{\frac{1}{3n} + 1} &\quad \text{for $n$ even},\\ \frac{2}{3}\frac{-1}{\frac{1}{3n} + 1} &\quad \text{for $n$ odd}.\end{cases}\] Note that for \(j \in \mathbb{N}\), \(a_{2j-1} \leq 0 \leq a_{2j}\). Also note that \((a_{2j-1})_j\) is a decreasing sequence and \((a_{2j})_{j}\) is an increasing sequence [Try showing these!] Moreover, \(\lvert a_n \rvert \leq \frac{2}{3} \; \forall n\in\mathbb{N}\), so \((a_n)_n\) is bounded.
Now, fix \(k \in \mathbb{N}\). We have: \[\begin{align*} \sup_{k\geq n}a_n &= \sup_{2j \geq k} a_{2j}, \; \; &&\text{(since only even elements are non-negative.)}\\ &= \lim_{j \to \infty} a_{2j}, \; \; &&\text{(since $(a_{2j})_j$ is a bounded increasing sequence)},\\ &= \frac{2}{3}. \; \; \quad &&\text{(by algebra of limits)} \end{align*}\] Hence, taking \(k \to \infty\), we find that \(\sup_{n \geq k} a_n \to \frac{2}{3}\). So, \(\limsup_{n \to \infty} a_n = \frac{2}{3}\).
Similarly, fixing \(k \in \mathbb{N}\) again: \[\begin{align*} \inf_{k\geq n}a_n &= \inf_{2j \geq k} a_{2j-1}, \; \; &&\text{(since only odd elements are non-positive.)}\\ &= \lim_{j \to \infty} a_{2j-1}, \; \; &&\text{(since $(a_{2j-1})_j$ is a bounded decreasing sequence)},\\ &= \lim_{j \to \infty}-\frac{4-\frac{2}{j}}{6 - \frac{2}{j}}, \; \; \quad &&\text{(by sequence definition)}\\ &= -\frac{2}{3}. \; \; \quad &&\text{(by algebra of limits)}. \end{align*}\]
Hence, taking \(k \to \infty\), we find that \(\inf_{n \geq k} a_n \to -\frac{2}{3}\). So, \(\liminf_{n \to \infty} a_n = -\frac{2}{3}\).