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编写小组/讲义/是的，这是一份礼物.md

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 是的，这是一份礼物
-![[image_editor_1761626924475.jpg]]
+
+---
+
+> 在实数系中，有界的单调数列必有极限。
+> （1）单调递增有上界的数列，必有极限且$\mathop {\lim }\limits_{n \to \infty } {a_n} = \sup \{ {a_n}\} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} (n \in N)$；
+> （2）单调递减有下界的数列，必有极限且$\mathop {\lim }\limits_{n \to \infty } {a_n} = \inf \{ {a_n}\} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} (n \in N)$.
+
+先来一道习题练练手吧~
+
+>[!example]  例一
+>设$a > 0,\sigma  > 0,{a_1} = \frac{1}{2}\left( {a + \frac{\sigma }{a}} \right),{a_{n + 1}} = \frac{1}{2}\left( {{a_n} + \frac{\sigma }{{{a_n}}}} \right),n = 1,2,...$
+> 证明：数列$\left\{ {{a_n}} \right\}$收敛，且极限为$\sqrt \sigma$.
+
+证：对 $\forall t > 0,t + \frac{1}{t} \ge 2$，令 $t = \frac{{{a_n}}}{{\sqrt \sigma  }} > 0$，则有${a_{n + 1}} = \frac{1}{2}\left( {{a_n} + \frac{\sigma }{{{a_n}}}} \right) = \frac{{\sqrt \sigma  }}{2}\left( {\frac{{{a_n}}}{{\sqrt \sigma  }} + \frac{{\sqrt \sigma  }}{{{a_n}}}} \right) \ge \frac{{\sqrt \sigma  }}{2}*2 = \sqrt \sigma     ,n = 1,2,...$ 
+故$\left\{ {{a_n}} \right\}$有下界
+又因为${a_{n + 1}} = \frac{1}{2}\left( {{a_n} + \frac{\sigma }{{{a_n}}}} \right) = \frac{{{a_n}}}{2}\left( {1 + \frac{\sigma }{{{a_n}^2}}} \right) \le \frac{{{a_n}}}{2}\left( {1 + \frac{\sigma }{\sigma }} \right) = {a_n},n = 1,2,...$
+所以$\left\{ {{a_n}} \right\}$单调递减.
+因此，由单调有界定理知:$\left\{ {{a_n}} \right\}$存在极限，即$\left\{ {{a_n}} \right\}$收敛.
+设$\mathop {\lim }\limits_{n \to \infty } {a_n} = A$，由${a_n} > 0$ 知 $A \ge 0$
+由${a_{n + 1}} = \frac{1}{2}\left( {{a_n} + \frac{\sigma }{{{a_n}}}} \right)$ 知 $2{a_{n + 1}}{a_n} = {a_n}^2 + \sigma$.
+两边令 $n \to \infty$取极限得: $2{A^2} = {A^2} + \sigma$
+解得$A =  - \sqrt \sigma$（舍）或$A =   \sqrt \sigma$.
+故$\mathop {\lim }\limits_{n \to \infty } {a_n} = \sqrt \sigma$.
+
+
+---
+
+
+>[!example] 例二  
+>证明：数列${x_n} = 1 + \frac{1}{2} + ... + \frac{1}{n} - \ln {\kern 1pt} n{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} (n = 1,2,...)\\$单调递减有界，从而有极限（此极限称为Euler常数，下记作$C$）。
+
+
+
+证法1️⃣：
+利用不等式$$\frac{1}{{1 + n}} < \ln (1 + \frac{1}{n}) < \frac{1}{n}$$有$${x_{n + 1}} - {x_n} = \frac{1}{{1 + n}} - \ln (n + 1) + \ln {\kern 1pt} n = \frac{1}{{1 + n}} - \ln (1 + \frac{1}{n}) < 0$$ 
+故数列${x_n}$严格递减.
+又因为
+$\begin{array}{l} {x_n} = \sum\limits_{k = 1}^n {\frac{1}{k}}  - \ln \left( {\frac{n}{{n - 1}} \cdot \frac{{n - 1}}{{n - 2}} \cdot ... \cdot \frac{3}{2} \cdot \frac{2}{1}} \right)\\   {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt}  = \sum\limits_{k = 1}^n {\frac{1}{k}}  - \sum\limits_{k = 1}^{n - 1} {\ln \left( {1 + \frac{1}{k}} \right)} \\  {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt}  = \sum\limits_{k = 1}^{n - 1} {\left[ {\frac{1}{k} - \ln \left( {1 + \frac{1}{k}} \right)} \right]}  + \frac{1}{n}\\  {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt}  > \frac{1}{n} > 0 \end{array}$
+所以数列${x_n}$有下界.
+因此，数列${x_n}$单调递减有下界，故$\mathop {\lim }\limits_{n \to \infty } {x_n}$存在.
+
+
+证法2️⃣：
+因为$$\left| {{x_n} - {x_{n + 1}}} \right| = \left| {\frac{1}{n} - \left[ {\ln {\kern 1pt} n - \ln (n - 1)} \right]} \right|$$
+对$\ln {\kern 1pt} n - \ln (n - 1)$利用Lagrange中值公式，有
+$$\ln {\kern 1pt} n - \ln (n - 1) = \frac{1}{{{\xi _n}}}，其中n - 1 < {\xi _n} < n.$$因此有$\left| {{x_n} - {x_{n - 1}}} \right| = \frac{{n - {\xi _n}}}{{n \cdot {\xi _n}}} < \frac{1}{{{{(n - 1)}^2}}}$
+而$\sum\limits_{n = 1}^\infty  {\frac{1}{{{{(n - 1)}^2}}}}$收敛，故$\sum\limits_{n = 1}^\infty  {\left| {{x_n} - {x_{n - 1}}} \right|}$收敛，从而${x_n} = \sum\limits_{k = 1}^n {\left( {{x_k} - {x_{k - 1}}} \right)}  + {x_1}$也收敛.
+因此，数列$\ {x_n}$收敛，即 $\mathop {\lim }\limits_{n \to \infty } {x_n}$ 存在.