高等数学笔记之：前奏

妖狐艹你老母 2021-09-25 10:42 334阅读 0赞

### 目录 ###

*  引入
 *  问题

# 引入 #

先来几道题，感受一下高等数学的魅力。

# 问题 #

问题 ①：  
 lim ⁡ n → ∞ ( 1 n 2 + n + 1 + 2 n 2 + n + 2 + ⋯ + n n 2 + n + n ) = ? \\lim\_\{n \\rightarrow \\infty\} \\left( \\frac\{1\}\{n^2+n+1\}+\\frac\{2\}\{n^2+n+2\}+\\dots+\\frac\{n\}\{n^2+n+n\} \\right)=? n→∞lim(n2\+n\+11\+n2\+n\+22\+⋯\+n2\+n\+nn)=?  
解答 ①：  
令  
 x n = ( 1 n 2 + n + 1 + 2 n 2 + n + 2 + ⋯ + n n 2 + n + n ) , x\_n = \\left( \\frac\{1\}\{n^2+n+1\}+\\frac\{2\}\{n^2+n+2\}+\\dots+\\frac\{n\}\{n^2+n+n\} \\right), xn=(n2\+n\+11\+n2\+n\+22\+⋯\+n2\+n\+nn),  
则有  
 ( 1 n 2 + n + n + 2 n 2 + n + n + ⋯ + n n 2 + n + n ) ≤ x n ( 1 + n ) n 2 n 2 + n + n ≤ x n ( 1 + n ) n 2 n ( n + 2 ) ≤ x n 1 2 ≤ n + 1 2 ( n + 2 ) ≤ x n , \\begin\{aligned\} \\left( \\frac\{1\}\{n^2+n+n\}+\\frac\{2\}\{n^2+n+n\}+\\dots+\\frac\{n\}\{n^2+n+n\} \\right) & \\leq x\_n \\\\ \\frac\{\\frac\{(1+n)n\}\{2\}\}\{n^2+n+n\} & \\leq x\_n \\\\ \\frac\{(1+n)n\}\{2n(n+2)\} & \\leq x\_n \\\\ \\frac\{1\}\{2\} \\leq \\frac\{n+1\}\{2(n+2)\} & \\leq x\_n, \\\\ \\end\{aligned\} (n2\+n\+n1\+n2\+n\+n2\+⋯\+n2\+n\+nn)n2\+n\+n2(1\+n)n2n(n\+2)(1\+n)n21≤2(n\+2)n\+1≤xn≤xn≤xn≤xn,

x n ≤ ( 1 n 2 + n + 1 + 2 n 2 + n + 1 + ⋯ + n n 2 + n + 1 ) x n ≤ ( 1 + n ) n 2 n 2 + n + 1 x n ≤ ( n 2 + n ) 2 ( n 2 + n + 1 ) ≤ 1 2 , \\begin\{aligned\} x\_n & \\leq \\left( \\frac\{1\}\{n^2+n+1\}+\\frac\{2\}\{n^2+n+1\}+\\dots+\\frac\{n\}\{n^2+n+1\} \\right) \\\\ x\_n & \\leq \\frac\{\\frac\{(1+n)n\}\{2\}\}\{n^2+n+1\} \\\\ x\_n & \\leq \\frac\{(n^2+n)\}\{2(n^2+n+1)\} \\leq \\frac\{1\}\{2\}, \\\\ \\end\{aligned\} xnxnxn≤(n2\+n\+11\+n2\+n\+12\+⋯\+n2\+n\+1n)≤n2\+n\+12(1\+n)n≤2(n2\+n\+1)(n2\+n)≤21,

故  
 1 2 ≤ x n ≤ 1 2 . \\frac\{1\}\{2\} \\leq x\_n \\leq \\frac\{1\}\{2\}. 21≤xn≤21.  
得到  
 lim ⁡ n → ∞ x n = lim ⁡ n → ∞ ( 1 n 2 + n + 1 + 2 n 2 + n + 2 + ⋯ + n n 2 + n + n ) = 1 2 . \\lim\_\{n \\rightarrow \\infty\} x\_n = \\lim\_\{n \\rightarrow \\infty\} \\left( \\frac\{1\}\{n^2+n+1\}+\\frac\{2\}\{n^2+n+2\}+\\dots+\\frac\{n\}\{n^2+n+n\} \\right) = \\frac\{1\}\{2\}. n→∞limxn=n→∞lim(n2\+n\+11\+n2\+n\+22\+⋯\+n2\+n\+nn)=21.

问题 ②：  
设  x n = 2 2 … 2 ⏟ n 重 根 号 x\_n = \\underbrace\{\\sqrt\{2\\sqrt\{2\\dots\\sqrt\{2\}\}\}\}\_\{n 重根号\} xn=n重根号22…2，考虑  \{ x n \} \\\{x\_n\\\} \{ xn\} 的收敛性，并求极限.

解答 ②：  
分析：  x 1 = 2 x\_1 = \\sqrt\{2\} x1=2， x 2 = 2 2 = 2 x 1 x\_2 = \\sqrt\{2\\sqrt\{2\}\} = \\sqrt\{2x\_1\} x2=22=2x1， … \\dots …， x n + 1 = 2 x n x\_\{n+1\} = \\sqrt\{2x\_n\} xn\+1=2xn（ x n > 0 x\_n > 0 xn>0）。  
假定  x k < 2 x\_k < 2 xk<2，则  x k + 1 = 2 x k < 2 ⋅ 2 = 2 x\_\{k+1\} = \\sqrt\{2x\_k\} < \\sqrt\{2 \\cdot 2\} = 2 xk\+1=2xk<2⋅2=2，  
所以  0 < x n < 2 0 < x\_n < 2 0<xn<2。  
 x n + 1 − x n = 2 x n − x n > x n ⋅ x n − x n = 0. x\_\{n+1\} - x\_n = \\sqrt\{2x\_n\} - x\_n > \\sqrt\{x\_n \\cdot x\_n\} - x\_n = 0. xn\+1−xn=2xn−xn>xn⋅xn−xn=0.  
 \{ x n \} \\\{x\_n\\\} \{ xn\} 单增有上界。

设  
 lim ⁡ n → ∞ x n = a , \\lim\_\{n \\rightarrow \\infty\} x\_n = a, n→∞limxn=a,  
则  
 a = 2 a , a = \\sqrt\{2a\}, a=2a,

得到  a = 2 a = 2 a=2。