单调有界数列必有极限的证明---改编自中科大数分教材

定理：单调有界数列必有极限

证明：仅证明单调递增有界数列必有极限，单调递减数列类似。
设{\(a_{n}\)}为单调递增数列，且有上界。
把该数列各项用十进制无限小数形式表示如下：
\(\quad\quad\quad\quad\quad\quad\)\(a_{1}=A_{1}.b_{11}b_{12}b_{13}\)......
\(\quad\quad\quad\quad\quad\quad\)\(a_{2}=A_{2}.b_{21}b_{22}b_{23}\)......
\(\quad\quad\quad\quad\quad\quad\)\(a_{3}=A_{3}.b_{31}b_{32}b_{33}\)......
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)..........
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)..........
其中\(A_{1},A_{2},A_{3}...是整数部分，b_{ij}是小数部分的数字\)，为0-9中的数字。\(\\\)
\(因为数列{a_{n}}是有界数列，所以整数部分不会无限增大，又因为{a_{n}}是有界数列，\)\(\\\)
\(所以整数数列A_{n}在达到最大值之后将固定不变，记该最大整数为A，并设A在第N_{0}行\)\(\\\)
\(上出现。考察第二列，即各行数字的小数部分的第一列b_{11}, b_{21}, b_{31}...\)\(\\\)
\(考察第A_{N_{0}}行和以下各行。\)\(\\\)
\(设x_{1}是出现在该列的最大数字，设其第一次出现在第N_{1}行上，其中N_{1}\geqslant N_{0}\).\(\\\)
\(则x_{1}一旦出现，不会改变，因为{a_{n}}是递增数列，且该行数字的整数部分，已经是最大值A\)。\(\\\)
\(继续考察各行数字的小数部分的第二列b_{21}, b_{31}, b_{41}...\)
\(同理可知，该列将出现一个最大固定值x_{2}，假设出现在第N_{2}行，这里N_{2}\geqslant N_{1}\)\(\\\)
\(\quad\quad\quad\quad\quad\quad\)\(a_{1}=A_{1}.b_{11}b_{12}b_{13}\)......
\(\quad\quad\quad\quad\quad\quad\)\(a_{2}=A_{2}.b_{21}b_{22}b_{23}\)......
\(\quad\quad\quad\quad\quad\quad\)\(a_{3}=A_{3}.b_{31}b_{32}b_{33}\)......
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(第N_{0}行:\quad\quad\)\(a_{N_{0}}=A.b_{N_{0}1}b_{N_{0}2}b_{N_{0}3}\)......
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(第N_{1}行:\quad\quad\)\(a_{N_{1}}=A.x_{1}b_{N_{1}2}b_{N_{1}3}\)......
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(第N_{2}行:\quad\quad\)\(a_{N_{2}}=A.x_{1}x_{2}b_{N_{2}3}\)......
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
重复以上过程，得到\(\quad\quad\)a=\(A.x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}\)......
\(下面证明，上式的a就是数列{a_{n}}的极限\)
\(\forall\varepsilon\)>0，设m\(\in\)\(N^{+}\),使得\(10^{-m}<\varepsilon\)
则有\(a_{N_{m}}\)=A.\(x_{1}x_{2}x_{3}\)......\(x_{m}b_{N_{m},m+1}\)......
注：\(a_{N_{m}}\)是指整数部分为A，小数部分的前m-1位，均为各自列的最大值\(x_{1}\)-\(x_{m-1}\)，\(\\\)
\(\quad\quad\)\(b_{n,m}\)这一列最大数字\(x_{m}\)第一次出现的那个数字\(\\\)
那么，\(\forall\)n>\(N_{m}\), \(A_{n}\)=A.\(x_{1}x_{2}x_{3}\)......\(x_{m}b_{n,m+1}\)......
a与\(a_{n}\) 的整数部分和前m位小数完全一样
|a-\(a_{n}\)|=|A.\(x_{1}x_{2}x_{3}\)......\(x_{m}x_{m+1}\)
\(\quad\quad\)-A.\(x_{1}x_{2}x_{3}\)......\(x_{m}b_{n,m+1}\)|<\(\frac{1}{10^{m}}\)<\(\varepsilon\)
所以 \(\lim_{n\to\infty}a_{n}\) = \(A.x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}\)......
即 \(\lim_{n\to\infty}a_{n} = a\)
证毕.