Definition 20.1
Definition 20.1. Let \(\{a_n\}\) be a bounded real sequence and let \(\mathcal{L}_a\) denote the set of all \(L\) such that \[ L = \lim_{k \to \infty} a_{n_k} \] where \(\{a_{n_k}\}\) is a convergent subsequence of \(\{a_n\}\). We define \[ \limsup_{n \to \infty} a_n = \sup \mathcal{L}_a \] and \[ \liminf_{n \to \infty} a_n = \inf \mathcal{L}_a \] The notations \(\overline{\lim}_{n \to \infty} a_n\) and \(\underline{\lim}_{n \to \infty} a_n\) are also used for \(\limsup_{n \to \infty} a_n\) and \(\liminf_{n \to \infty} a_n\), respectively.
Remark
Remark. Let $\{ a_n \}$ be a bounded sequence by $M$. Then
(1) There is a convergent subsequence of $\{ a_n \}$ by Bolzano-Weierstrass Theorem, which means that $\mathcal{L}_a$ is not empty.
(2) Since $\mathcal{L}_a$ is bounded by $M$ (($\because$) We have $|a_n| \leq M, \forall n \in \mathbb{P}$ and $\lim_{k \to \infty} |a_{n_k}| \leq M $by Theorem 14.2.), there exists the least upper bound and the greatest lower bound by A13 and Theorem 5.4.
Theorem 20.2
Theorem 20.2. Let \(\{a_n\}\) be a bounded sequence. Then \[ \liminf_{n \to \infty} a_n \leq \limsup_{n \to \infty} a_n. \]
Proof. Let $\{ a_{n_k} \}$ be a subseqence of $\{ a_n \}$ such that $\lim_{n \to \infty} a_{n_k} = L$ (by Bolzano-Weierstrass Theorem). Then $$\liminf_{n \to \infty} a_n = \inf \mathcal{L}_a \leq L \leq \sup \mathcal{L}_a = \limsup_{n \to \infty} a_n$$ by Definition 20.1. Thus $$ \liminf_{n \to \infty} a_n \leq \limsup_{n \to \infty} a_n. \blacksquare$$
Theorem 20.3
Theorem 20.3. Let \(\{a_n\}\) be a bounded sequence, and let \(L = \limsup_{n \to \infty} a_n\) and \(M = \liminf_{n \to \infty} a_n\).
(i) If \(\varepsilon > 0\), there exist infinitely many positive integers \(n\) such that \(L - \varepsilon < a_n\), and there exists a positive integer \(N_1\) such that if \(n \geq N_1\), then \(a_n < L + \varepsilon\).
(ii) If \(\varepsilon > 0\), there exist infinitely many positive integers \(n\) such that \(a_n < M + \varepsilon\), and there exists a positive integer \(N_2\) such that if \(n \geq N_2\), \(M - \varepsilon < a_n\).
Proof. Suppose that it is false that $L - \varepsilon < a_n$ for infinitely many positive integers $n$. Then $L - \varepsilon < a_n$ for finitely many positive integers $n$. Let $N_1$ be the maximum among such positive integers. Then $a_n \leq L - \varepsilon, \forall n > N_1$. Then $\{ a_n \}_{n=N_1 + 1}^{\infty}$ is a subsequence of $\{ a_n \}$ and it is also bounded. By Bolzano-Weierstarss Theorem, $\{ a_n \}_{n = N_1 + 1}^{\infty}$ has a convergent subsequence $\{ a_{n_k} \}$. Then $a_{n_k} \leq L - \varepsilon, \forall k \in \mathbb{P}$. By Theorem 14.2, $\lim_{k \to \infty} a_{n_k} \leq L - \varepsilon$, which means that $L - \varepsilon$ is an upper bound for $\mathcal{L}_a$. Thus $L \leq L - \varepsilon. \bigotimes$. Therefore there exist infinitely many positive integers \(n\) such that \(L - \varepsilon < a_n\).
Suppose that it is false that there exists $N_1 \in \mathbb{P}$ such that $a_n < L + \varepsilon, \forall n \geq N_1$. Then $L + \varepsilon \leq a_n$ for infinitely many positive integers $n$. Let $\{ a_{n_k} \}$ be a subsequence of $\{ a_n \}$ such that $L + \varepsilon \leq a_{n_k}$ for all positive integers $k$. Since it is also bounded, there is a convergent subsequence $\{ a_{n_{k_j}} \}$ of $\{ a_{n_k} \}$. Then $L + \varepsilon \leq a_{n_{k_j}}, \forall j \in \mathbb{P}$, and by Theorem 14.2, $L + \varepsilon \leq \lim_{j \to \infty} a_{n_{k_j}} \leq L$, which is a contradiction. Therefore there exists a positive integer \(N_1\) such that if \(n \geq N_1\), then \(a_n < L + \varepsilon\). $\blacksquare$
직관적으로 이해해보자. $L$은 bounded sequence $\{ a_n \}$의 극한값의 upper bound 중 최소값이다. 그러면 $L$ 보다 작거나 같은 값으로 수렴하는 수많은 부분수열들이 있을 거고, 수열의 극한의 정의에 의해 $L - \varepsilon$ 보다 큰 $a_n$들이 존재하는 인덱스 $N_1, N_2, ...$를 생각할 수 있다. 그런데 일반적으로 이런 $N_1, N_2, ...$들 중 max가 항상 존재하지는 않고, 따라서 특정 인덱스 이후로 무한히 많다고 말할 수 없으므로 $L - \varepsilon < a_n$을 만족하는 자연수는 just 무한히 많이 있다.
그러나 $L + \varepsilon > a_n$을 만족하는 자연수에 대해서는 반드시 특정 인덱스 이후로 무한히 많다고 말할 수 있다. 만약 그렇지 않다면, 다시 말해 특정 인덱스 이상의 자연수 $n$에 대해 $L + \varepsilon$ 보다 크거나 같은 $a_n$이 존재한다고 해보자. 만약 이런 $n$이 유한개 있다면 별 문제가 되지 않는다. 그냥 유한 개의 자연수 이후로 인덱스를 잡아버리면 된다. 그런데 이런 $n$이 무한개 있다고 해보자. 그러면 이런 $n$들만 모아놓은 새로운 수열을 구성해 낼 수 있는데, 기존 수열이 bounded 이므로 이런 수열도 bounded이고, 따라서 Bolzano-Weierstrass theorem에 의해 수렴하는 부분수열이 존재한다. 그런데 이렇게 잡아낸 부분수열의 극한값은 $L$보다 크다. 이는 모순이므로 이러한 상황은 불가능하다.
Theorem 20.4
Theorem 20.4. (i) Let \(\{a_n\}\) be a sequence such that \[ \lim_{n \to \infty} a_n = L \] Then \[ \limsup_{n \to \infty} a_n = L = \liminf_{n \to \infty} a_n. \]
(ii) Let \(\{a_n\}\) be a bounded sequence such that \[ \limsup_{n \to \infty} a_n = L = \liminf_{n \to \infty} a_n \] Then \[ \lim_{n \to \infty} a_n = L. \]
Proof. (i) Since $\lim_{n \to \infty} a_n = L$, $\{ a_n \}$ is bounded and every convergent subsequence converges to $L$, which means that $\mathcal{L}_a = \{ L \}$. Then $$ \limsup_{n \to \infty} a_n = L = \liminf_{n \to \infty} a_n. $$ (ii) Let $\varepsilon > 0$. By Theorem 20.3, $\exists N_1 \in \mathbb{P}$ such that $a_n < L + \varepsilon, \forall n \geq N_1$, and $\exists N_2 \in \mathbb{P}$ such that $L - \varepsilon < a_n, \forall n \geq N_2$. Then $$L - \varepsilon < a_n < L + \varepsilon, \forall n \geq \max \{ N_1, N_2 \},$$ which means that $\lim_{n \to \infty} a_n = L.$ $\blacksquare$
Theorem 20.5
Theorem 20.5. Let \(\{a_n\}\) and \(\{b_n\}\) be bounded sequences such that \(a_n \leq b_n\) for every positive integer \(n\). Then \[ \limsup_{n \to \infty} a_n \leq \limsup_{n \to \infty} b_n \] and \[ \liminf_{n \to \infty} a_n \leq \liminf_{n \to \infty} b_n. \]
Proof. Let $\{ a_{n, k} \}$ be a subseqence of $\{ a_n \}$ with the limit $L$. Then $a_{n, k} \leq b_{n, k}, \forall k \in \mathbb{P}$. Since $\{ b_n \}$ is bounded, $\{ b_{n, k} \}$ is also bounded, which means that it has a convergent subsequence $\{ b_{n, k, j} \}$. Then $$L = \lim_{n \to \infty} a_{n, k} = \lim_{j \to \infty} a_{n, k, j} \leq \lim_{j \to \infty} b_{n, k, j} \leq \limsup_{n \to \infty} b_n.$$ Thus $\limsup_{n \to \infty} b_n$ is an upper bound for $\mathcal{L}_a$, which means that $$\limsup_{n \to \infty} a_n \leq \limsup_{n \to \infty} b_n. \blacksquare$$
Theorem 20.6
Theorem 20.6. Let \(\{a_n\}\) and \(\{b_n\}\) be bounded sequences. Then \[ \limsup_{n \to \infty} (a_n + b_n) \leq \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n \] and \[ \liminf_{n \to \infty} a_n + \liminf_{n \to \infty} b_n \leq \liminf_{n \to \infty} (a_n + b_n). \]
Proof. Let $\{ a_{n, k} + b_{n, k} \}$ be a subsequence of $\{ a_n + b_n \}$ with the limit $L$. Then $\{ a_{n, k} \}$ and $\{ b_{n, k} \}$ is subsequences of $\{ a_n \}$ and $\{ b_n \}$, respectively.
Let $\{ a_{n, k, j} \}$ be a convergent subseqence of $\{ a_{n, k} \}$. Then $\{ b_{n, k, j} \}$ is a subseqence of $\{ b_{n, k} \}$. Since $\{ b_n \}$ is bounded, so is $\{ b_{n, k, j} \}$. Then it has a convergent subsequence $\{ b_{n, k, j, l} \}$, and $\{ a_{n, k, j, l} \}$ is a subsequence of $\{ a_{n, k} \}$. Thus we have $$L = \lim_{k \to \infty} (a_{n, k} + b_{n, k}) = \lim_{k \to \infty} a_{n, k} + \lim_{k \to \infty} b_{n, k} \\ = \lim_{l \to \infty} a_{n, k, j, l} + \lim_{l \to \infty} b_{n, k, j, l} \leq \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n.$$ Then $\limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n$ is an upper bound for $\mathcal{L}_{a+b}$ which is the set of the limits of subsequences of $\{ a_n + b_n \}$. Thus $$\limsup_{n \to \infty} (a_n + b_n) \leq \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n \blacksquare$$
Theorem 20.7
Theorem 20.7. Let \(\{a_n\}\) be a sequence such that \[ \lim_{n \to \infty} a_n = L \] Then \[ \lim_{n \to \infty} \frac{a_1 + a_2 + \cdots + a_n}{n} = L. \]
Proof.
Theorem 20.8
Theorem 20.8. Let \(\{a_n\}\) and \(\{b_n\}\) be sequences such that \[ \lim_{n \to \infty} a_n \geq 0 \] and \(\{b_n\}\) is a bounded sequence. Then \[ \limsup_{n \to \infty} a_n b_n = \limsup_{n \to \infty} a_n \limsup_{n \to \infty} b_n = \lim_{n \to \infty} a_n \limsup_{n \to \infty} b_n. \]
Proof. Since $\{ a_n \}$ is convergent, $$\limsup_{n \to \infty} a_n \limsup_{n \to \infty} b_n = \lim_{n \to \infty} a_n \limsup_{n \to \infty} b_n.$$ Let $\lim_{n \to \infty} a_n = L$. If $L = 0$, then $\lim_{n \to \infty} a_nb_n = 0$ by Theorem 13.3, and $$\limsup_{n \to \infty} a_nb_n = 0 = \limsup_{n \to \infty} a_n \limsup_{n \to \infty} b_n.$$ Suppose $L > 0$. Since $\{ b_n \}$ is bounded, there is a convergent subsequence $\{ b_{n_k} \}$ of $\{ b_n \}$. Then we have $$\lim_{k \to \infty} a_{n_k} b_{n_k} = \lim_{k \to \infty} a_{n_k} \lim_{k \to \infty} b_{n_k} = \limsup_{n \to \infty} a_n \lim_{k \to \infty} b_{n_k} \leq \limsup_{n \to \infty} a_n \limsup_{n \to \infty} b_n.$$ Thus $\limsup_{n \to \infty} a_n \limsup_{n \to \infty} b_n$ is an upper bound for $\lim_{k \to \infty} a_{n_k} b_{n_k}$, and we have $$\limsup_{n \to \infty} a_n b_n \leq \limsup_{n \to \infty} a_n \limsup_{n \to \infty} b_n.$$ Furthermore, $$\limsup_{n \to \infty} a_n \lim_{k \to \infty} b_{n_k} = \lim_{k \to \infty} a_{n_k} b_{n_k} \leq \limsup_{n \to \infty} a_n b_n \\ \Longrightarrow \lim_{k \to \infty} b_{n_k} \leq \frac{\limsup_{n \to \infty} a_n b_n}{\limsup_{n \to \infty} a_n},$$ which means that $$\frac{\limsup_{n \to \infty} a_n b_n}{\limsup_{n \to \infty} a_n}$$ is an upper bound for $\lim_{k \to \infty} b_{n_k}$. Then $$\limsup_{n \to \infty} b_n \leq \frac{\limsup_{n \to \infty} a_n b_n}{\limsup_{n \to \infty} a_n} \\ \Longrightarrow \limsup_{n \to \infty} a_n \limsup_{n \to \infty} b_n \leq \limsup_{n \to \infty} a_b b_n.$$ Thus we have $$\limsup_{n \to \infty} a_n b_n = \limsup_{n \to \infty} a_n \limsup_{n \to \infty} b_n. \blacksquare$$