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单调有界收敛定理及其应用
weierstrass(魏尔斯特拉斯)定理
定理 2.4.1:
单调增的数列\(\{u_n\}\)有极限\(\iff\)\(\{u_n\}\)有上界
单调减的数列\(\{u_n\}\)有极限\(\iff\)\(\{u_n\}\)有下界
证明:
"\(\Rightarrow\)": 显然
"\(\Leftarrow\)": 只证\(\{u_n\}\)单增有上界时 \(\lim\limits_{k\to \infty}{u_n}\)存在:
由确界原理知, \(\{u_n\}\)有上确界\(\beta\), 下证\(\lim\limits_{n\to\infty}u_n = \beta\) \[
\beta = \sup\{u_n\} \Rightarrow \forall n \in \mathbb{N^+}, u_n \le
\beta \tag{1}
\] \[
\forall \varepsilon > 0, \exists N \in \mathbb{N^+}, s.t. u_N >
\beta - \varepsilon \tag{2}
\] \[
\{u_n\}单增 \Rightarrow n > N时, u_n \ge u_N \tag{3}
\] 由\((1)(2)(3)\)可知 \[
\forall \varepsilon > 0, \exists N\in \mathbb{N^+}, 当n > N时,
\beta - \varepsilon < u_n \le u_n \le \beta < \beta + \varepsilon
\] \[
即\lim\limits_{n\to\infty} u_n = \beta = \sup\{u_n\}
\]
EX. 2.4.1: 若\(x_1 = 1, x_{n + 1} = \frac{1}{2}(x_n + \frac{2}{x_n}), n\in\mathbb{N^+}\), 证明\(\{x_n\}\)收敛, 并求出\(\lim\limits_{n \to \infty}x_n\)
证明:
step. 1: 证\(\{x_n\}\)有下界(归纳法)
用归纳法证明\(x_n > \sqrt 2, \forall n >
1, n \in \mathbb{N^+}\):
\(n = 2\)时\(x_2 = \frac{3}{2} > 2\)成立
假设\(n = k\)时成立, 下证\(n = k + 1\)时成立
\[
x_{k + 1} = \frac{1}{2}(x_{k} + \frac{2}{k_k})\ge\frac{1}{2} \cdot 2
\cdot \sqrt{x_k \cdot \frac{2}{x_k}} = \sqrt{2}
\] \[
``="成立\iff x_k = \frac{2}{x_k} 即 x_k = \sqrt{2}
\] \[
\therefore \forall n > 2, n\in\mathbb{N^+}, x_n > \sqrt{2}
\]
step. 2: 证\(\{x_n\}\)单减
\[ x^2 > 2(x_n > 0) \Rightarrow \frac{2}{x_n} < x_n \Rightarrow x_{n + 1} = \frac{1}{2}(x_n + \frac{2}{x_n}) < x_n \] \[ \therefore \{x_n\}单减 \]
由定理2.4.1可知, \(\lim\limits_{n \to \infty}{x_n}\)存在, 记为\(a\), 在\(x_{n + 1} = \frac{1}{2}(x_n + \frac{2}{x_n})\)两边令\(n\to \infty\) \[ a = \frac{1}{2}(a + \frac{2}{a}) \] \[ a = \sqrt{2} \]
注: 在证明\(\{x_n\}\)单调时,
一般采用下列方法:
1. 讨论\(x_n - x_{n + 1}\)
如在EX. 2.4.1中, \(x_{n + 1} = \frac{1}{2}(x_n
+ \frac{2}{x_n})\), \(x_n =
\frac{1}{2}(x_{n - 1} + \frac{2}{x_{n - 1}})\)
\[
\begin{aligned}
x_{n + 1} - x_{n} &= \frac{1}{2}(x_n - x_{n - 1}) +
(\frac{1}{x_n} - \frac{1}{x_{n - 1}})\\
&=(x_n - x_{n + 1})(\frac{1}{2} - \frac{1}{x_n\cdot x_{n - 1}})
\end{aligned}\] \[
{x_n}单调\iff x_n + 1 - x_n 与 x_n - x_{n + 1}同号
\] 2. 讨论\(\frac{x_n}{x_{n +
1}}\)
EX. 2.4.2: \(x_1 > 0, x_{n + 1} = 1 +
\frac{x_n}{1 + x_n}, \forall n\in\mathbb{N^+}\), 证明\(\lim\limits_{n\to\infty}x_n\)存在并求出
证明:
step. 1: 证明 \(1 < x_n < 2, \forall n
\ge 2, n\in\mathbb{N^+}\)
这里的 \(x_n < 2\) 是用 \(a = 1 + \frac{a}{1 + a}\) 猜出来的
运用归纳法: \(n = 2\) 时, \(x_2 = 1 + \frac{x_1}{1 + x_1} \in (1,
2)\)
假设当 \(n = k\) 时, \(x_n \in (1, 2)\) , 所以\(x_{k + 1} = 1 + \frac{x_k}{1 + x_k} \in (1,
2)\)
step. 2: 证明\(\{x_n\}\)单调 \[ x_{n + 1} - x_n = \frac{x_n - x_{n - 1}}{(1 + x_n)(1 + x_{n - 1})} \] \[ (1 + x_n)(1 + x_{n - 1}) > 0 \Rightarrow x_{n + 1} - {x_n}与x_n - x_{n - 1}同号 \] \[ \Rightarrow \{x_n\}单调 \Rightarrow \lim\limits_{n\to\infty}x_n 存在, 记为a \]
step. 3: 求\(a\)
在 \(x_{n + 1} = 1 + \frac{x_n}{1 +
x_n}\) 两边令 \(n\to\infty\)
\[ a = 1 + \frac{a}{1 + a} (a \in [1, 2]) \] \[ \Rightarrow a = \frac{1 + \sqrt{5}}{2} \]
EX. 2.4.3: 证明 \(\lim\limits_{n\to\infty}\frac{a^n}{n!} = 0\,\, (a > 1)\)
证明: 令 \(u_n = \frac{a^n}{n!} > 0\), 则 \(\frac{u_{n + 1}}{u_n} = \frac{a}{n + 1} \to 0 < 1\)
\[
\therefore\exists N\in\mathbb{N^+}, n > N 时, \frac{u_{n + 1}}{u_n}
< 1, 即 u_{n + 1} < u_n
\] \(u_n > 0\),
根据单调有界原理, \(\lim\limits_{n\to\infty}u_n\)存在,
记为\(A \ge 0\)
由 \(x_n + 1 = \frac{a ^ {n + 1}}{(n + 1)!} =
x_n \cdot \frac{a}{n + 1}\), 令两边\(n\to\infty\), 则 \[
\begin{aligned}
A &= 0 \cdot A\\
A &= 0
\end{aligned}
\]
单调有界原理的应用: \(\pi\) 和 \(e\)
\(\pi\) : 割圆术
单位圆内有一内接正\(n\)边形,
边长为\(a\), 则 \[
\begin{aligned}
a^2 &= 4\sin^2\frac{180^\circ}{n}\\
a &= 2\sin\frac{180^\circ}{n}
\end{aligned}
\] 所以正\(n\)边形的半周长 \(L_n = \frac{n}{2}\cdot a = n \cdot
\sin\frac{180^\circ}{n}\)
下证