What follows introduces the concept of the limit superior and the limit inferior of sequences.
Consider \(x_{n} = (-1)^{n}\). Note that \(x_{2n} = 1\) and \(x_{2n+1} = -1\). In particular,
\[\begin{aligned} \lim_{ n \to \infty } x_{2n} &=1, \\ \lim_{ n \to \infty } x_{2n+1} &= -1. \end{aligned}\]Consider
\[\begin{aligned} a_{n} &= \underset{m \in \mathbb{N}}{\inf} \{ x_{m}:m\geq n \} \\ b_{n} &= \underset{m \in \mathbb{N}}{\sup} \{ x_{m}:m\geq n \}. \end{aligned}\]Note that \(a_{n} \leq x_{n} \leq b_{n}\). Moreover, for every \(k \in \mathbb{N}\), \(a_{n} \leq x_{n+k} \leq b_{n}\).
Limits superior and inferior
Let \(\{x_{n}\}_{n=1}^\infty \subseteq \mathbb{R}\). Define
\[\begin{aligned} a_{n} &= \underset{m \in \mathbb{N}}{\inf} \{ x_{m}:m\geq n \} \\ b_{n} &= \underset{m \in \mathbb{N}}{\sup} \{ x_{m}:m\geq n \}. \end{aligned}\]Note that \(a_{n} \leq x_{n+k} \leq b_{n}\) for every \(k \in \mathbb{N}\). Moreover, note that
\[a_{n} = \inf \{ x_{m}: m\geq n \} \leq \inf \{ x_{m}:m\geq n+1 \} = a_{n+1},\] \[b_{n+1} = \sup \{ x_{m}:m\geq n+1 \} \leq \sup \{ x_{m}:m\geq n \} = b_{n}.\]Assume that the sequence is bounded, that is, that there exist \(N,M \in \mathbb{R}\) such that \(N \leq x_{n} \leq M\) for every \(n \in \mathbb{N}\). Then \(N \leq \inf \{ x_{m}:m\geq n \} = a_{n}\) and \(b_{n} = \sup \{ x_{m}:m\geq n \} \leq M\).
Definition
Let \(\{x_{n}\}_{n=1}^\infty \subseteq \mathbb{R}\) be bounded. We define
\[\begin{aligned} \limsup x_{n} &= \lim_{ n \to \infty } b_{n} \\ \liminf x_{n} &= \lim_{ n \to \infty } a_{n}. \end{aligned}\]Since \(b_{n+1}\leq b_{n}\), note that, by the monotone convergence theorem,
\[\limsup x_{n} =\lim_{ n \to \infty } b_{n} = \underset{n}{\inf} b_{n} = \underset{n}{\inf} \underset{m}{\sup} \{ x_{m} : m\geq n\}.\]In the same way,
\[\liminf x_{n} = \lim_{ n \to \infty } a_{n} = \underset{n}{\sup} a_{n} = \underset{n}{\sup} \underset{m}{\inf} \{ x_{m}: m\geq n \}.\]Lemma (Bounds on limits of subsequences)
Let \(\{x_{n}\}_{n=1}^\infty\) be a bounded sequence and let \(\{x_{n}\}_{n=1}^\infty\) have a subsequence \(\{x_{k_{n}}\}_{n=1}^\infty\) converging to \(L.\)Then \(\liminf x_{n} \leq L \leq \limsup x_{n}\).
Proof: Consider a subsequence \(\{x_{k_{n}}\}_{n=1}^\infty\). Note that \(x_{k_{n}} \leq b_{n}\) since \(k_{n} \geq n\). In the same way, \(a_{n} \leq x_{k_{n}}\). Hence \(a_{n} \leq x_{k_{n}} \leq b_{n}\). The result follows on taking limits.
Theorem (Convergence and the limits superior and inferior)
Let \(\{x_{n}\}_{n=1}^\infty\) be a sequence. Then \(\lim_{ n \to \infty } x_{n} = L\) if and only if \(\limsup x_{n} = \liminf x_{n} = L\).
Proof: (\(\implies\)): Let \(\{x_{n}\}_{n=1}^\infty\) be such that \(x_{n} \underset{n \rightarrow \infty}{\longrightarrow}L\). Then \(\{x_{n}\}_{n=1}^\infty\) is bounded. Hence there exist subsequences \(\{x_{k_{n}}\}_{n=1}^\infty\) and \(\{x_{\ell_{n}}\}_{n=1}^\infty\) such that \(x_{k_{n}} \underset{n \rightarrow \infty}{\longrightarrow} \liminf x_{n}\) and \(x_{\ell_{n}} \underset{n \rightarrow \infty}{\longrightarrow} \limsup x_{n}\) (exercise). Conclude that \(\limsup x_{n} = \liminf x_{n} = L\), since every subsequence must converge to \(L\). (\(\impliedby\)): Let \(\{x_{k_{n}}\}_{n=1}^\infty\) be a fixed and arbitrary subsequence. We know that \(a_{n} \leq x_{k_{n}} \leq b_{n}\). Then, by the squeeze theorem, conclude that \(x_{k_{n}} \underset{n \rightarrow \infty}{\longrightarrow} L\). Since every subsequence converges to \(L\) (because we took a fixed and arbitrary one), conclude that \(x_{n} \underset{n \rightarrow \infty}{\longrightarrow} L\).
A very common trick is to use the definitions of \(\limsup\) and \(\liminf\), and to note that:
\[\inf x_{n} \leq \\\liminf x_{n} \leq \limsup x_{n} \leq \sup x_{n}.\]If \(m\) is a lower bound of \(x_{n}\) for every \(n\geq k\) and \(M\) is an upper bound of \(y_{n}\) for every \(n \geq k\), then by the definitions of \(\sup y_{n}\) and \(\inf x_{n}\) we have that \(m \leq \underset{n\geq k}{\inf} x_{n}\) and that \(M \geq \underset{n\geq k}{\sup} y_{n}\).