Stolz 的应用

例题 1

对于数列 $x_0=a,0<a<{\displaystyle \frac{\pi }{2}},x_n=\sin x_{n-1}(n=1,2,\cdots )$ ，证明:

$$\underset{n\to \mathrm{\infty }}{lim}{x}_n=0,\underset{n\to \mathrm{\infty }}{lim}\sqrt{\frac{n}{3}}{x}_n=1.$$

解答

因为 $0<a<{\displaystyle \frac{\pi }{2}},x_0=a$ ，递推可知

$$0<x_n=\sin x_{n-1}<x_{n-1}<{\displaystyle \frac{\pi }{2}}(n=1,2,\cdots )$$

$\{x_n\}$ 单调递减且有下界 $0$ ， $\underset{n\to \mathrm{\infty }}{lim}{x}_n$ 存在. 记 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{x}_n=A}$ ，知 $A=\sin A\Rightarrow A=0$ ， ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{x}_n=0.}$

要证 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sqrt{\frac{n}{3}}{x}_n=1}$ ，即证 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{n}{{\displaystyle \frac{1}{x_n^2}}}=3}$

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n}{{\displaystyle \frac{1}{x_n^2}}}}& \stackrel{\text{Stolz}}{=}\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n-(n-1)}{{\displaystyle \frac{1}{x_n^2}}-{\displaystyle \frac{1}{x_{n-1}^2}}}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{{\displaystyle \frac{1}{{\sin }^2{x}_{n-1}}}-{\displaystyle \frac{1}{x_{n-1}^2}}}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{x_{n-1}^2{\sin }^2{x}_n-1}{x_{n-1}^2-{\sin }^2{x}_{n-1}}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{x^2{\sin }^{2}x}{x^2-{\sin }^{2}x}}=\underset{x\to 0}{lim}{\displaystyle \frac{x^4}{(x+\sin x)(x-\sin x)}}=\underset{x\to 0}{lim}{\displaystyle \frac{x^4}{(2x+o\left(x\right))({\displaystyle \frac{x3}{6}}+o\left(x^3\right))}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{1}{(2+o\left(1\right))({\displaystyle \frac{1}{6}}+o\left(1\right))}}=3.\end{array}$$

得证 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sqrt{\frac{n}{3}}{x}_n=1.}$

例题 2

设 $0<a_1<1,a_{n+1}=a_n(1-a_n)(\mathrm{\forall }n\in \mathbb{N})$ ，证明: ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}n{a}_n=1.}$

解答

由 $0<x_1<1$ 及 $x_2=x_1(1-x_1)$ 知， $0<x_2<1$ ，用数学归纳法可证: $\mathrm{\forall }n\in {\mathbb{N}}^{\ast }:0<x_n<1$ ，于是 $0<{\displaystyle \frac{x_{n+1}}{x_n}}=1-x_n<1(n=1,2,\cdots )$ ，
从而 $\{x_n\}\searrow 0$ ，不妨设 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{x}_n=A}$ ，递推关系式两边取极限，得 $A=A(1-A)$ ，解得 $A=0.$
令 $b_n={\displaystyle \frac{1}{x_n}}$ ，则 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{b}_n=+\mathrm{\infty }}$ ，且数列 $\{b_n\}$ 是严格单调递增，故由 Stolz 定理

$${\displaystyle \underset{n\to \mathrm{\infty }}{lim}n{x}_n=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n}{{\displaystyle \frac{1}{x_n}}}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n}{b_n}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{b_{n+1}-b_n}}=\underset{n\to \mathrm{\infty }}{lim}(1-x_n)=1.}$$

例题 3

设 $x_1>0,x_{n+1}=\ln (1+x_n)(n=1,2,\cdots )$ ，求 $\underset{n\to \mathrm{\infty }}{lim}n{x}_n.$

解答

$x_2=\ln (1+x_1)>0$ ，用数学归纳法可证 $\mathrm{\forall }n\in {\mathbb{N}}^{\ast }:x_n>0$ ，又 $x_1>0,x_{n+1}=\ln (1+x_n)<x_n$ ，故数列 $\{x_n\}\searrow 0$ ，那么

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}n{x}_n& =\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n}{{\displaystyle \frac{1}{x_n}}}}\stackrel{\text{Stolz}}{=}\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{{\displaystyle \frac{1}{x_n}}-{\displaystyle \frac{1}{x_n-1}}}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{{\displaystyle \frac{1}{\ln (1+x_{n-1})}}-{\displaystyle \frac{1}{x_{n-1}}}}}\\ & =\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{x_{n-1}\ln (1+x_{n-1})}{x_{n-1}-\ln (1+x_{n-1})}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{x_{n-1}^2}{{\displaystyle \frac{1}{2}}{x}_{n-1}^2}}=2.\end{array}$$

例题 4

序列 $a_{ij}={\displaystyle \frac{i+j}{i^2+j^2}}$ ，求极限 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{n}}\sum _{i=1}^n\sum _{j=1}^n{a}_{ij}.}$

解答

由 Stolz ( $\ast /\mathrm{\infty }$ 型) 得 (以下的括号不为矩阵符号)

$$\begin{array}{rl}& \left(\begin{array}{cccccccccc}& {\displaystyle \frac{1+1}{1^2+1^2}}& +& {\displaystyle \frac{1+2}{1^2+2^2}}& +& \cdots & +& {\displaystyle \frac{1+n}{1^2+n^2}}& +& {\displaystyle \frac{1+n+1}{1^2+(n+1{)}^2}}\\ +& {\displaystyle \frac{2+1}{2^2+1^2}}& +& {\displaystyle \frac{2+2}{2^2+2^2}}& +& \cdots & +& {\displaystyle \frac{2+n}{2^2+n^2}}& +& {\displaystyle \frac{2+n+1}{2^2+(n+1{)}^2}}\\ ⋮& ⋮& & ⋮& & & & ⋮& & ⋮\\ +& {\displaystyle \frac{n+1}{n^2+1^2}}& +& {\displaystyle \frac{n+2}{n^2+2^2}}& +& \cdots & +& {\displaystyle \frac{n+n}{n^2+n^2}}& +& {\displaystyle \frac{n+n+1}{n^2+(n+1{)}^2}}\\ +& {\displaystyle \frac{n+1+1}{(n+1{)}^2+1^2}}& +& {\displaystyle \frac{n+1+2}{(n+1{)}^2+2^2}}& +& \cdots & +& {\displaystyle \frac{n+1+n}{(n+1{)}^2+n^2}}& +& {\displaystyle \frac{n+1+n+1}{(n+1{)}^2+(n+1{)}^2}}\end{array}\right)\\ & -\left(\begin{array}{cccccccc}& {\displaystyle \frac{1+1}{1^2+1^2}}& +& {\displaystyle \frac{1+2}{1^2+2^2}}& +& \cdots & +& {\displaystyle \frac{1+n}{1^2+n^2}}\\ +& {\displaystyle \frac{2+1}{2^2+1^2}}& +& {\displaystyle \frac{2+2}{2^2+2^2}}& +& \cdots & +& {\displaystyle \frac{2+n}{2^2+n^2}}\\ ⋮& ⋮& & ⋮& & & & ⋮\\ +& {\displaystyle \frac{n+1}{n^2+1^2}}& +& {\displaystyle \frac{n+2}{n^2+2^2}}& +& \cdots & +& {\displaystyle \frac{n+n}{n^2+n^2}}\end{array}\right)=2\left[\sum _{k=1}^n{\displaystyle \frac{(n+1)+k}{(n+1{)}^2+k^2}}\right]+{\displaystyle \frac{1}{n+1}}.\end{array}$$ $$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{n}}\sum _{i=1}^n\sum _{j=1}^n{\displaystyle \frac{i+j}{i^2+j^2}}& =\underset{n\to \mathrm{\infty }}{lim}(\sum _{i=1}^{n+1}\sum _{j=1}^{n+1}-\sum _{i=1}^n\sum _{j=1}^n){\displaystyle \frac{i+j}{i^2+j^2}}=\underset{n\to \mathrm{\infty }}{lim}[2\left(\sum _{k=1}^n{\displaystyle \frac{(n+1)+k}{(n+1{)}^2+k^2}}\right)+{\displaystyle \frac{1}{n+1}}]\\ & =\underset{n\to \mathrm{\infty }}{lim}[{\displaystyle \frac{2}{n+1}}\left(\sum _{k=1}^n{\displaystyle \frac{1+{\displaystyle \frac{k}{n+1}}}{1+{\left({\displaystyle \frac{k}{n+1}}\right)}^2}}\right)+{\displaystyle \frac{1}{n+1}}]=2{\int }_0^1{\displaystyle \frac{1+x}{1+x^2}}\mathrm{d}x={\displaystyle \frac{\pi }{2}}+\ln 2.\end{array}$$