Example 2
Solution First, note that for any \(j\in\mathbb{N}\), \(a_{2j} \leq 2 \leq a_{2j-1}\). But this time, we find that both \((a_{2j})_j\) and \((a_{2j-1})_j\) are decreasing sequences! In this case, the argument used in Example 1 will only work for \(\liminf_{n\to\infty} a_n.\) Try using that argument to show that \[\liminf_{n\to\infty}a_n = 1.\] For \(\limsup_{n\to\infty} a_n\), we have to look towards the start of the sequence. To this end, fix \(k \in \mathbb{N}\). Then, \[\begin{align*} \sup_{n\geq k}a_n &= \sup_{2j-1 \geq k} a_{2j - 1} \; \; &&\text{(since only the odd elements are $\geq 2$)}\\ &=\begin{cases} a_k \; \text{if $k$ is odd},\\ a_{k+1} \; \text{if $k$ is even}\end{cases} \; \; &&\text{(because $(a_{2j-1})_j$ is a decreasing sequence)}\\ &=\begin{cases} \frac{1}{k^2} + 3 \; \text{if $k$ is odd},\\ \frac{1}{(k+1)^2} + 3 \; \text{if $k$ is even}\end{cases} \end{align*}\]
In both cases, as \(k \to \infty\), \(\sup_{n\geq k }a_n \to 3\), so \(\limsup_{n \to \infty} a_n = 3\).