Sequence
Definition 10.1. Let $X$ be a set. A sequence of elements of $X$ is a function from $\mathbb{P}$ into $X$.
특별히 real sequence라고 하면 $X = \mathbb{R}$인 경우로, $\{ a_n \}^{\infty}_{n = 1}$ 혹은 $\{ a_n \}$으로 쓴다. $a_n$을 단독으로 쓰면 $n$th term이라고 부른다.
Limit of a Sequence
Definition 10.2. Let $\{a_n\}$ be a sequence of real numbers. We say that has $\{a_n\}$ has limit $L \in \mathbb{R}$ if for every $\varepsilon > 0$, there exists a positive integer $N$, such that if $n \geq N$, then $$|a_n - L| < \varepsilon .$$
반대로 어떤 수열 $\{ a_n \}$이 수렴하지 않으면 다음의 명제가 성립한다.
Remark. A sequence $\{a_n\}$ does not have limit $L$ if for some $\varepsilon > 0$, $|a_n − L| < \varepsilon$ for infinitely many positive integers $n$.
Theorem 10.3
Theorem 10.3. The limit of a sequence is unique. That is, if a sequence $\{a_n\}$ has limit $L$ and limit $L′$, then $L = L′$.
Proof. Suppose that $L > L'$. Let $\varepsilon = \frac{|L - L'|}{2}$. Then there exists $N_1$ and $N_2$ such that $|a_n - L| < \frac{|L - L'|}{2}$ for all $n \geq N_1$ and $|a_n - L'| < \frac{|L - L'|}{2}$ for all $n \geq N_2$. Let $N = \max \{N_1, N_2 \}$. Then $|L - L'| = |L - a_n + a_n - L'| \leq |a_n - L| + |a_n - L'| < |L - L'| \bigotimes$. Similarly, we can show that if $L < L'$, then we have a contradiction. Thus $L = L'$. $\blacksquare$
Theorem 12.1
Theorem 12.1. If $a_n = L$ for all $n \in \mathbb{P}$, then $$\lim_{n \to \infty} a_n = L.$$
Proof. Let $\varepsilon > 0$. Then for any $n \geq 1$, $|a_n - L| = 0 < \varepsilon$. Thus $$\lim_{n \to \infty} a_n = L. \blacksquare$$
Theorem 12.2
Theorem 12.2. Let $\{a_n\}$ and $\{b_n\}$ be sequences such that $\lim_{n \to \infty} a_n = L$ and $\lim_{n \to \infty} b_n = M$. Then $$\lim_{n \to \infty} (a_n + b_n) = L+M.$$
Proof. Let $\varepsilon > 0$. Then $\exists N_1, N_2 \in \mathbb{P}$ such that $|a_n - L| < \frac{\varepsilon}{2}$ for all $n \geq N_1$ and $|b_n - M| < \frac{\varepsilon}{2}$ for all $n \geq N_2$. Let $N = \max \{N_1, N_2 \}$. Then $|(a_n + b_n) - (L + M)| = |a_n - L + b_n - M| \leq |a_n - L| + |b_n - M| < \varepsilon$ for all $n \geq N$. Thus $$\lim_{n \to \infty} (a_n + b_n) = L + M. \blacksquare$$
Theorem 12.3
Theorem 12.3. Let $\{a_n\}$ be a sequence such that $\lim_{n \to \infty} a_n = L$. If $c$ is any real number, then $$\lim_{n \to \infty} ca_n = cL.$$
Proof. Let $\varepsilon > 0$. Then $\exists N \in \mathbb{P}$ such that $|a_n - L| < \frac{\varepsilon}{c}$ for all $n \geq N$. Thus $|ca_n - cL| = c|a_n - L| < c \cdot \frac{\varepsilon}{c} = \varepsilon, \forall n \geq N$, so $$\lim_{n \to \infty} ca_n = cL. \blacksquare$$
Corollary 12.4
Corollary 12.4. Let $\{a_n\}$ and $\{b_n\}$ be sequences such that $\lim_{n \to \infty} a_n = L$ and $\lim_{n \to \infty} b_n = M$. Then $$\lim_{n \to \infty} (a_n - b_n) = L - M.$$
Proof. By Theorem 12.2 and Theorem 12.3, the result is clear. $\blacksquare$
Lemma 12.5
Lemma 12.5. Let $\{a_n\}$ and $\{b_n\}$ be sequences such that $$\lim_{n \to \infty} a_n = 0 = \lim_{n \to \infty} b_n.$$ Then $\lim_{n \to \infty} a_n b_n = 0$.
Proof. Let $\varepsilon > 0$. Then $\exists N_1, N_2 \in \mathbb{P}$ such that $|a_n| < \sqrt{\varepsilon}$ for all $n \geq N_1$ and $|b_n| < \sqrt{\varepsilon}$ for all $n \geq N_2$. Let $N = \max \{N_1, N_2 \}$. Then $|a_n b_n - 0| = |a_n| |b_n| < \varepsilon$ for all $n \geq N$. Then $$\lim_{n \to \infty} a_n b_n = 0.$$
Theorem 12.6
Theorem 12.6. Let $\{a_n\}$ and $\{b_n\}$ be sequences such that $\lim_{n \to \infty} a_n = L$ and $lim_{n \to \infty} b_n = M$. Then $\lim_{n \to \infty} a_n b_n = LM$.
Proof. Note that $\lim_{n \to \infty} (a_n - L) = 0$ and $\lim_{n \to \infty} (b_n - M) = 0$ by Theorem 12.1 and Corollary 12.4. Then $$\lim_{n \to \infty} (a_n - L)(b_n - M) = 0$$ by Lemma 12.5 and $$\lim_{n \to \infty} a_nM = LM = \lim_{n \to \infty} b_nL$$ by Theorem 12.3. Then $$\lim_{n \to \infty} a_n b_n
= \lim_{n \to \infty} \left[ (a_n - L)(b_n - M) + a_n M + L b_n - L M \right] \\ = \lim_{n \to \infty} (a_n - L)(b_n - M) + \lim_{n \to \infty} a_n M
+ \lim_{n \to \infty} L b_n + \lim_{n \to \infty} (-L M) \\ = 0 + L M + L M - L M = L M. \blacksquare$$
Corollary 12.7
Corollary 12.7. Let $\{a_n\}$ be a sequence such that $\lim_{n \to \infty} a_n = L$, and let $k$ be a positive integer. Then $\lim_{n \to \infty} a^k_n = L^k$.
Proof. By Theorem 12.6 and mathematical induction, we can draw the conclusion. $\blacksquare$
Lemma 12.8
Lemma 12.8. Let $\{a_n\}$ be a sequence such that $\lim_{n \to \infty} a_n = L \neq 0$. Then $a_n \neq 0$ for all but finitely many positive integers $n$, and $$\lim_{n \to \infty} \frac{1}{a_n} = \frac{1}{L}.$$
Proof. Let $\varepsilon = \frac{|L|}{2}$. Then $\exists N_1 \in \mathbb{P}$ such that $|a_n - L| < \frac{|L|}{2}$ for all $n \geq N_1$. Then $$|L| = |L - a_n + a_n| \leq |a_n - L| + |a_n| < \frac{|L|}{2} + |a_n|, \forall n \geq N_1 \\ \Longrightarrow \frac{|L|}{2} \leq |a_n|, \forall n \geq N_1.$$ Thus $a_n \neq 0$ for all $n \geq N_1$.
Let $\varepsilon > 0$. Then $\exists N_2 \in \mathbb{P}$ such that $|a_n - L| < \frac{|L|^2 \varepsilon}{2}$ for all $n \geq N_2$. If we let $N = \max \{ N_1, N_2 \}$, then we have $$\Bigg|\frac{1}{a_n} - \frac{1}{L} \Bigg| = \Bigg| \frac{L - a_n}{a_nL} \Bigg| = \frac{|a_n - L|}{|a_n| |L|} < \frac{2}{|L|} \frac{1}{|L|} \frac{|L|^2 \varepsilon}{2} = \varepsilon, \forall n \geq N.$$ Thus $\lim_{n \to \infty} \frac{1}{a_n} = \frac{1}{L}.$ $\blacksquare$
Theorem 12.9
Theorem 12.9. Let $\{a_n\}$ and $\{b_n\}$ be sequences such that $\lim_{n \to \infty} a_n = L$ and $\lim_{n \to \infty} b_n = M \neq 0$. Then $$\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{L}{M}.$$
Proof. By Theorem 12.6 and Lemma 12.8, $$\lim_{n \to \infty} a_n \frac{1}{b_n} = \lim_{n \to \infty} a_n \lim_{n \to \infty} \frac{1}{b_n} = L \frac{1}{M}. \blacksquare$$
Lemma 14.1
Lemma 14.1. Let $\{a_n\}$ be a sequence such that $\ leq a_n$ for all $n \in \mathbb{P}$ and $L = \lim_{n \to \infty} a_n$. Then $0 \leq L$.
Proof. Suppose that $L < 0$. Let $\varepsilon = \frac{|L|}{2} = - \frac{L}{2}$. Then $\exists N \in \mathbb{P}$ such that $$|a_n - L| < -\frac{L}{2} \\ \Longrightarrow 0 \leq a_n < \frac{L}{2} < 0 \bigotimes$$ Thus $0 \leq L$. $\blacksquare$
Theorem 14.2
Theorem 14.2. Let $\{a_n\}$ and $\{b_n\}$ be sequences such that \[ \lim_{n \to \infty} a_n = L \quad \text{and} \quad \lim_{n \to \infty} b_n = M. \] If $a_n \leq b_n$ for every positive integer $n$, then $L \leq M$.
Proof. Let $c_n = b_n - a_n$. By Corollary 12.4 and Lemma 14.1, $L \leq M$. $\blacksquare$
Squeeze Theorem
Theorem 14.3 (Squeeze Theorem). Let $\{a_n\}$, $\{b_n\}$, and $\{c_n\}$ be sequences such that \[ a_n \leq b_n \leq c_n \] for every positive integer $n$. If \[ \lim_{n \to \infty} a_n = L = \lim_{n \to \infty} c_n, \] then \[ \lim_{n \to \infty} b_n = L. \]
Proof. Let $\varepsilon > 0$. Then $\exists N_1, N_2 \in \mathbb{P}$ such that $|a_n - L| < \varepsilon, \forall n \geq N_1$ and $|c_n - L|, \forall n \geq N_2$. Let $N = \max \{N_1, N_2\}$. Then $$L - \varepsilon < a_n \leq b_N \leq c_n < L+ \varepsilon, \forall n \geq N.$$ Thus $\lim_{n \to \infty} b_n = L$. $\blacksquare$
Exercise 10.12
Exercise 10.12. Let $\{ a_n \}$ be a sequence such that $\lim_{n \to \infty} a_n = L$. Then $\lim_{n \to \infty} |a_n| = |L|$.
Solution. Since $\lim_{n \to \infty} a_n = L$, $\forall \varepsilon > 0, \exists N$ such that $|a_n - L| < \varepsilon, \forall n \geq N$. Note that for all $n \geq N$, $$ |a_n| = |a_n - L + L| \leq |a_n - L| + |L| < |L| + \varepsilon \\ \Longrightarrow |a_n| - |L| < \varepsilon$$ and $$|L| = |L - a_n + a_n| \leq |a_n - L| + |a_n| < |a_n| + \varepsilon \\ \Longrightarrow - \varepsilon < |a_n| - |L|.$$ Thus we have that for all $n \geq N$, $$-\varepsilon < |a_n| - |L| < \varepsilon \\ \Longrightarrow | |a_n| - |L| | < \varepsilon \\ \Longrightarrow \lim_{n \to \infty} |a_n| = |L|. \blacksquare$$