Stolz 的应用

例题 1

对于数列 \(x_0=a,0<a<\dfrac{\pi}{2},x_n=\sin x_{n-1}~ (n=1,2,\cdots)\)，证明: \[\lim_{n\to\infty}x_n=0,~\lim_{n\to\infty}\sqrt{\frac{n}{3}}x_n=1.\]

解答

因为 \(0<a<\dfrac{\pi}{2},~x_0=a\)，递推可知 \[0<x_n=\sin x_{n-1}<x_{n-1}<\dfrac{\pi}{2}~ (n=1,2,\cdots)\]

\(\{x_n\}\) 单调递减且有下界 \(0\)，\(\lim\limits_{n\to\infty}x_n\) 存在. 记 \(\displaystyle\lim_{n\to\infty}x_n=A\)，知 \(A=\sin A\Rightarrow A=0\)，\(\displaystyle\lim_{n\to\infty}x_n=0.\)

要证 \(\displaystyle \lim_{n\to\infty}\sqrt{\frac{n}{3}}x_n=1\)，即证 \(\displaystyle\lim_{n\to\infty}\frac{n}{\dfrac{1}{x_n^2}}=3\)

\[\begin{align*} \lim _{n\rightarrow \infty }\dfrac{n}{\dfrac{1}{x_{n}^{2}}} & \xlongequal[]{\text{Stolz}}\lim _{n\rightarrow \infty }\dfrac{n-\left( n-1\right) }{\dfrac{1}{x_{n}^{2}}-\dfrac{1}{x_{n-1}^{2}}}=\lim _{n\rightarrow \infty }\dfrac{1}{\dfrac{1}{\sin ^{2}x_{n-1}}-\dfrac{1}{x_{n-1}^{2}}}=\lim _{n\rightarrow \infty }\dfrac{x_{n-1}^{2}\sin ^{2}x_{n}-1}{x_{n-1}^{2}-\sin ^{2}x_{n-1}} \\ & =\lim _{x\rightarrow 0}\dfrac{x^{2}\sin ^{2}x}{x^{2}-\sin ^{2}x}=\lim _{x\rightarrow 0}\dfrac{x^{4}}{\left( x+\sin x\right) \left( x-\sin x\right) }=\lim _{x\rightarrow 0}\dfrac{x^{4}}{\left( 2x+o\left( x\right) \right) \left( \dfrac{x3}{6}+o\left( x^{3}\right) \right) } \\ & =\lim _{x\rightarrow 0}\dfrac{1}{\left( 2+o\left( 1\right) \right) \left( \dfrac{1}{6}+o\left( 1\right) \right) }=3. \end{align*}\]

得证 \(\displaystyle \lim_{n\to\infty}\sqrt{\frac{n}{3}}x_n=1.\)

例题 2

设 \(0<a_1<1,a_{n+1}=a_n(1-a_n)~ (\forall n\in\mathbb{N})\)，证明: \(\displaystyle\lim_{n\to\infty}na_n=1.\)

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: \(\forall n\in\mathbb{N}^*:0<x_n<1\)，于是 \(0<\dfrac{x_{n+1}}{x_{n}}=1-x_n<1~ (n=1,2,\cdots)\)， 从而 \(\{x_n\}\searrow 0\)，不妨设 \(\displaystyle\lim_{n\to\infty}x_n=A\)，递推关系式两边取极限，得 \(A=A(1-A)\)，解得 \(A=0.\) 令 \(b_n=\dfrac{1}{x_n}\)，则 \(\displaystyle\lim_{n\to\infty}b_n=+\infty\)，且数列 \(\{b_n\}\) 是严格单调递增，故由 Stolz 定理

\[\displaystyle\lim _{n\rightarrow \infty }nx_{n}=\lim _{n\rightarrow \infty }\dfrac{n}{\dfrac{1}{x_{n}}}=\lim _{n\rightarrow \infty }\dfrac{n}{b_{n}}=\lim _{n\rightarrow \infty }\dfrac{1}{b_{n+1}-b_{n}}=\lim _{n\rightarrow \infty }\left( 1-x_{n}\right) =1.\]

例题 3

设 \(x_1>0,x_{n+1}=\ln(1+x_n)~ (n=1,2,\cdots)\)，求 \(\lim\limits_{n\to\infty}nx_n.\)

解答

\(x_2=\ln(1+x_1)>0\)，用数学归纳法可证 \(\forall n\in \mathbb{N}^*:x_n>0\)，又 \(x_1>0,x_{n+1}=\ln(1+x_n)<x_n\)，故数列 \(\{x_n\}\searrow 0\)，那么 \[\begin{align*} \lim _{n\rightarrow \infty }nx_{n} & =\lim _{n\rightarrow \infty }\dfrac{n}{\dfrac{1}{x_{n}}}\xlongequal[]{\text{Stolz}}\lim _{n\rightarrow \infty }\dfrac{1}{\dfrac{1}{x_{n}}-\dfrac{1}{x_{n}-1}}=\lim _{n\rightarrow \infty }\dfrac{1}{\dfrac{1}{\ln \left( 1+x_{n-1}\right) }-\dfrac{1}{x_{n-1}}} \\ & =\lim _{n\rightarrow \infty }\dfrac{x_{n-1}\ln \left( 1+x_{n-1}\right) }{x_{n-1}-\ln \left( 1+x_{n-1}\right) }=\lim _{n\rightarrow \infty }\dfrac{x_{n-1}^{2}}{\dfrac{1}{2}x_{n-1}^{2}}=2. \end{align*}\]

例题 4

序列 \(a_{ij}=\dfrac{i+j}{i^2+j^2}\)，求极限 \(\displaystyle\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}.\)

解答

由 Stolz (\(*/\infty\) 型) 得 (以下的括号不为矩阵符号) \[\begin{align*} & \begin{pmatrix} & \dfrac{1+1}{1^2+1^2} & + & \dfrac{1+2}{1^2+2^2} & + & \cdots & + & \dfrac{1+n}{1^2+n^2} & + & \dfrac{1+n+1}{1^2+(n+1)^2} \\ + & \dfrac{2+1}{2^2+1^2} & + & \dfrac{2+2}{2^2+2^2} & + & \cdots & + & \dfrac{2+n}{2^2+n^2} & + & \dfrac{2+n+1}{2^2+(n+1)^2} \\ \vdots & \vdots & & \vdots & & & & \vdots & & \vdots \\ + & \dfrac{n+1}{n^2+1^2} & + & \dfrac{n+2}{n^2+2^2} & + & \cdots & + & \dfrac{n+n}{n^2+n^2} & + & \dfrac{n+n+1}{n^2+(n+1)^2} \\ + & \dfrac{n+1+1}{(n+1)^2+1^2} & + & \dfrac{n+1+2}{(n+1)^2+2^2} & + & \cdots & + & \dfrac{n+1+n}{(n+1)^2+n^2} & + & \dfrac{n+1+n+1}{(n+1)^2+(n+1)^2} \end{pmatrix} \\ & - \begin{pmatrix} & \dfrac{1+1}{1^2+1^2} & + & \dfrac{1+2}{1^2+2^2} & + & \cdots & + & \dfrac{1+n}{1^2+n^2} \\ + & \dfrac{2+1}{2^2+1^2} & + & \dfrac{2+2}{2^2+2^2} & + & \cdots & + & \dfrac{2+n}{2^2+n^2} \\ \vdots & \vdots & & \vdots & & & & \vdots \\ + & \dfrac{n+1}{n^2+1^2} & + & \dfrac{n+2}{n^2+2^2} & + & \cdots & + & \dfrac{n+n}{n^2+n^2} \end{pmatrix} =2\left[ \sum ^{n}_{k=1}\dfrac{(n+1) +k}{(n+1) ^{2}+k^{2}}\right] +\dfrac{1}{n+1}. \end{align*}\] \[\begin{align*} \lim _{n\rightarrow \infty }\dfrac{1}{n}\sum ^{n}_{i=1}\sum ^{n}_{j=1}\dfrac{i+j}{i^{2}+j^{2}} & =\lim _{n\rightarrow \infty }\left( \sum ^{n+1}_{i=1}\sum ^{n+1}_{j=1}-\sum ^{n}_{i=1}\sum ^{n}_{j=1}\right) \dfrac{i+j}{i^{2}+j^{2}}=\lim _{n\rightarrow \infty }\left[ 2\left( \sum ^{n}_{k=1}\dfrac{(n+1) +k}{(n+1) ^{2}+k^{2}}\right) +\dfrac{1}{n+1}\right] \\ & =\lim _{n\rightarrow \infty }\left[ \dfrac{2}{n+1}\left( \sum ^{n}_{k=1}\dfrac{1+\dfrac{k}{n+1}}{1+\left( \dfrac{k}{n+1}\right) ^{2}}\right) +\dfrac{1}{n+1}\right] =2\int _{0}^{1}\dfrac{1+x}{1+x^{2}} \mathrm{d} x=\dfrac{\pi}{2}+\ln 2. \end{align*}\]