Monotone Sequences

2025. 5. 7. 10:20·Mathematics/Real analysis

Monotone Sequences

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 \[ \frac{b^{n+1} - a^{n+1}}{b - a} < (n+1) b^n. \]
Proof. Since $0 \leq a < b$, $$\frac{b^{n+1} - a^{n+1}}{b - a} = b^n + b^{n-1}a + \cdots + a^n < (n+1)b^n. \blacksquare$$

Theorem 16.6

Theorem 16.6. The sequence $\left\{ \left(1 + \frac{1}{n} \right)^n \right\}$ is increasing and convergent. The limit is denoted $e$.
Proof. By Lemma 16.5, for all $n \in \mathbb{P}$, $$\frac{\left( 1 + \frac{1}{n} \right)^{n+1} - \left( 1 + \frac{1}{n+1} \right)^{n+1}}{\left( 1 + \frac{1}{n} \right) - \left( 1 + \frac{1}{n+1} \right)} \\ = n(n+1) \left[ \left( 1 + \frac{1}{n} \right)^{n+1} - \left( 1 + \frac{1}{n+1} \right)^{n+1} \right] < (n+1) \left( 1 + \frac{1}{n} \right)^n \\ = (n+1) \left( \frac{n+1}{n} \right)^n \\ \Longrightarrow \left( 1 + \frac{1}{n} \right)^{n+1} - \left( 1 + \frac{1}{n+1} \right)^{n+1} < \frac{1}{n} \left( 1 + \frac{1}{n} \right)^n \\ \Longrightarrow \left( 1 + \frac{1}{n} \right)^n < \left( 1+ \frac{1}{n+1} \right)^{n+1}.$$ Thus $\left\{ \left(1 + \frac{1}{n} \right)^n \right\}$ is increasing. 
Note that $$\frac{\left( 1 + \frac{1}{2n} \right)^{n+1} - 1^{n+1}}{1 + \frac{1}{2n} - 1} = 2n \left( 1 + \frac{1}{2n} \right)^{n+1} - 1 < (n+1) \left( 1+ \frac{1}{2n} \right)^n \\ \Longrightarrow \left( 1 + \frac{1}{2n} \right)^{n+1} - 1 < \left( \frac{1}{2} + \frac{1}{2n} \right) \left( 1 + \frac{1}{2n} \right)^n \\ \Longrightarrow \frac{1}{2} \left( 1 + \frac{1}{2n} \right)^n - 1 \leq 0 \\ \Longrightarrow \left( 1 + \frac{1}{n} \right)^n < \left( 1 + \frac{1}{2n} \right)^{2n} \leq 4.$$ Thus $\left\{ \left(1 + \frac{1}{n} \right)^n \right\}$ is bounded. By Theorem 16.2, it is convergent. $\blacksquare$

Theorem 16.7

Theorem 16.7. The sequence $\left\{n^{1/n}\right\}_{n=3}^\infty$ is decreasing and \[ \lim_{n \to \infty} n^{1/n} = 1. \]
Proof. By the above proof, we have obtained $$\left( 1 + \frac{1}{n} \right)^n \leq 4, \forall n \in \mathbb{P}.$$ Note that $\left( 1 + \frac{1}{3} \right)^3 = \frac{64}{27} = 2 + \frac{10}{27} < 3$. Thus $$\left( 1 + \frac{1}{n} \right)^n \leq n, \forall n \geq 3.$$ Then for any $n \geq 3$, $$\left( 1 + \frac{1}{n} \right)^n \leq n \Longrightarrow (n+1)^n \leq n^{n+1} \\ \Longrightarrow (n+1)^{\frac{1}{n+1}} \leq n^{\frac{1}{n}}.$$ Thus $\{ n^{\frac{1}{n}} \}_{n=3}^{\infty}$ is decreasing. 
Since $1 \leq n^{\frac{1}{n}} \leq 3^{\frac{1}{3}}, \forall n \geq 3$, $\{ n^{\frac{1}{n}} \}$ is bounded. Then by Theorem 16.2, it is convergent.
Let $\lim_{n \to \infty} n^{\frac{1}{n}} = L$. Then $L^2 = \lim_{n \to \infty} n^{\frac{2}{n}}$. Note that $$\lim_{n \to \infty} \left( \frac{1}{2} \right)^{\frac{2}{n}} = \lim_{n \to \infty} \left( \frac{1}{4} \right)^{\frac{1}{n}} = 1$$ by Theorem 16.4. Then by Theorem 12.6, $$L^2 = L^2 \cdot 1 = \lim_{n \to \infty} n^{\frac{2}{n}} \cdot \left( \frac{1}{2} \right)^{\frac{2}{n}} = \lim_{n \to \infty} \left( \frac{n}{2} \right)^{\frac{2}{n}}.$$ Since $\{ n^{\frac{1}{n}} \}$ is a subsequence of $\left \{ \left( \frac{n}{2} \right)^{\frac{2}{n}} \right \}$, we have $L = L^2$, which implys that $L = 1$. $\blacksquare$
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