Definition 16.1. Let $\{a_n\}$ be a sequence. We say that $\{a_n\}$ is increasing (decreasing) if $a_n \leq a_{n+1}$ ($a_n \geq a_{n+1}$) for every positive integer $n$. We say that the sequence $\{a_n\}$ is monotone if either $\{a_n\}$ is increasing or $\{a_n\}$ is decreasing. If $a_n < a_{n+1}$ ($a_n > a_{n+1}$) for every positive integer $n$, we say that $\{a_n\}$ is strictly increasing (strictly decreasing). We say that $\{a_n\}$ is strictly monotone if either $\{a_n\}$ is strictly increasing or strictly decreasing.
Theorem 16.2
Theorem 16.2. A monotone sequence $\{a_n\}$ is convergent if and only if $\{a_n\}$ is bounded.
Proof. $(\Longrightarrow)$ We have proven this in Theorem 13.2. $(\Longleftarrow)$ Suppose that $\{a_n\}$ is increasing. Let $A = \{ a_n \,|\, n \in \mathbb{P} \}$. Since $\{ a_n \}$ is bounded above, so is $A$. Then by A13, there exists $a = \sup A$. Let $\varepsilon > 0$. Note that $a - \varepsilon$ is not an upper bound for $A$. (If it is, then $a < a - \varepsilon \Longrightarrow \varepsilon < 0 \bigotimes$) Then $\exists N \in \mathbb{P}$ such that $a - \varepsilon < a_N$. Since $\{a_n\}$ is increasing, $a - \varepsilon < a_n \leq a < a + \varepsilon, \forall n \geq N$. This means that $|a_n - a| < \varepsilon, \forall n \geq N$. Thus $\lim_{n \to \infty} a_n = a$. $\blacksquare$
Remark
Remark. By the above proof, if $\{ a_n \}$ is an increasing and bounded sequence, then the least upper bound for the set $A = \{ a_n \, | \, n \in \mathbb{P} \}$ is the limit of $\{ a_n \}$. On the other hand, the limit of $\{ a_n \}$ is the least upper bound for $A$.
Theorem 16.3
Theorem 16.3. If $|a| < 1$, then \[ \lim_{n \to \infty} a^n = 0. \]
Proof. Let $a_n = a^n, \forall n \in \mathbb{P}$. If $0 \leq a < 1$, then $1 > a > a^2 > a^3 > \cdots$, so $a^n = |a|^n = |a^n| < 1, \forall n \in \mathbb{P}$. Thus $\{a_n\}$ is bounded, and by Theorem 16.2, $\{a_n\}$ converges. Let $\lim_{n \to \infty} a_n = L$. Note that $\{ a_{n+1} \}$ is a subsequence of $\{a_n\}$. Then by Theorem 11.2 and Theorem 12.3, $$L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} a^{n+1} = \lim_{n \to \infty} a \cdot a^n = a \lim_{n \to \infty}a^n = aL \\ \Longrightarrow (a-1)L = 0 \\ \Longrightarrow L = 0$$ because $a \neq 1$. If $-1 < a < 0$, then $0 < -a < 1$. Thus $\lim_{n \to \infty} (-a)^n = 0$. Note that $\{(-1)^n\}$ is bounded. Then by Theorem 13.3, $$\lim_{n \to \infty} (-1)^n \cdot (-a)^n = \lim_{n \to \infty} (-1)^{2n} a^n = \lim_{n \to \infty} a^n = 0. \blacksquare$$
Theorem 16.4
Theorem 16.4. If $a > 0$, then \[ \lim_{n \to \infty} a^{1/n} = 1. \]
Proof. Suppose that $\{ a^{\frac{1}{n}} \}$ converges to some value $L$. Since $a^{\frac{1}{n+1}}$ is a subsequence of $a^{\frac{1}{n}}$, $a^{\frac{1}{n+1}}$ also converges to $L$. Then by Theorem 12.9, we have $$1 = \frac{L}{L} = \frac{\lim_{n \to \infty} a^{\frac{1}{n}}}{\lim_{n \to \infty} a^{\frac{1}{n+1}}} = \lim_{n \to \infty} a^{\frac{1}{n} - \frac{1}{n+1}} = \lim_{n \to \infty} a^{\frac{1}{n^2 + n}}. $$ Since $a^{\frac{1}{n^2 + n}}$ is a subsequence of $a^{\frac{1}{n}}$, it converges to $L$. Thus $L = 1$. Now, let's prove $\{ a^{\frac{1}{n}}\}$ is convergent. If $a \geq 1$, then $\{ a^{\frac{1}{n}} \}$ is decreasing because for all $n \in \mathbb{P}$, $$1 \leq a \Longrightarrow 1 \cdot a^n \leq a \cdot a^n \\ \Longrightarrow a^n \leq a^{n+1} \Longrightarrow a^{\frac{1}{n+1}} \leq a^{\frac{1}{n}}.$$ Note that $$1 < a^{\frac{1}{n}} \leq a, \forall n \in \mathbb{P}.$$ Thus $\{a^{\frac{1}{n}} \}$ is bounded. Then by Theorem 16.2, $\{a^{\frac{1}{n}}\}$ is convergent. If $0 < a < 1$, then $\{a^{\frac{1}{n}}\}$ is incresing because for all $n \in \mathbb{P}$, $$a < 1 \Longrightarrow a \cdot a^n = a^{n+1} < a^n \\ \Longrightarrow a^{\frac{1}{n}} < a^{\frac{1}{n+1}}.$$ Since $a^{\frac{1}{n}} < 1$, $\{a^{\frac{1}{n}}\}$ is bounded. Then by Theorem 16.2, it is convergent. $\blacksquare$
Lemma 16.5
Lemma 16.5. Let $a$ and $b$ be numbers such that $0 \leq a < b$. Then \[ \frac{b^{n+1} - a^{n+1}}{b - a} < (n+1) b^n. \]