考研数学非数竞赛复习之Stolz定理求解数列极限

2025-03-08 约 647 字预计阅读 2 分钟

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考研数学&非数竞赛复习之Stolz定理求解数列极限

在非数类大学生数学竞赛中，Stolz定理作为一种强大的工具，经常被用来解决和式数列极限的问题，也被誉为离散版的’洛必达’方法，它提供了一种简洁而有效的方法，使得原本复杂繁琐的极限计算过程变得直观明了。本文，我们将通过几个例题介绍该定理的使用方法。

stolz定理

1.设数列 $\left { a_n \right }$

, $\left { b_n \right }$

满足: $\left { b_n \right }$

严格单调递增

且 $\lim_{n\to\infty}\left { b_n \right }=\infty$

$\lim_{n\to\infty}\left { a_n \right }=\infty$

若 $https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Ba_n-a_n-1%7D%7Bb_n-b_n-1%7D%3DL$

则 $https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Ba_n%7D%7Bb_n%7D%3DL$

此为 $https://latex.csdn.net/eq?%5Cfrac%7B%5Cinfty%7D%7B%5Cinfty%7D$ 型未定式

2.设数列 $\left { a_n \right }$

, $\left { b_n \right }$

满足: $\left { b_n \right }$

严格单调递减

且 $\lim_{n\to\infty}\left { b_n \right }=0$

$\lim_{n\to\infty}\left { a_n \right }=0$

若 $https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Ba_n-a_n-1%7D%7Bb_n-b_n-1%7D%3DL$

则 $https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Ba_n%7D%7Bb_n%7D%3DL$

此为 $https://latex.csdn.net/eq?%5Cfrac%7B0%7D%7B0%7D$ 型未定式

定理看起来非常简单易懂,且该定理与洛必达公式形似。洛必达公式描述的是函数的导数的极限与原函数的极限之间的关系,该定理描述的是数列差分后的极限与原数列极限之间的关系。

例题

$https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto0%7D%5Cfrac%7B1+%5Csqrt%7B2%7D+%5Csqrt%5B3%5D%7B3%7D+...%5Csqrt%5Bn%5D%7Bn%7D%7D%7Bn%7D$ （ $https://latex.csdn.net/eq?%5Cfrac%7B%5Cinfty%7D%7B%5Cinfty%7D$ ）

解:设 $1+\sqrt{2}+\sqrt[3]{3}+…\sqrt[n]{n}$ ,

$a_n=\sum_{k=1}^{n}\sqrt[k]{k}$

$a_{n-1}=\sum_{k=1}^{n-1}\sqrt[k]{k}$

$a_n-a_{n-1}=\sqrt[n]{n}$

$b_n-b_{n-1}=n-(n-1)=1$

设 $https://latex.csdn.net/eq?L%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Ba_n-a_%7Bn-1%7D%7D%7Bb_n-b_%7Bn-1%7D%7D%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%7D%7D%7B1%7D$

则 $https://latex.csdn.net/eq?L%3D%5Clim_%7Bn%5Cto%5Cinfty%7De%5E%7B%5Cfrac%7B%5Cln%20n%7D%7Bn%7D%7D%3De%5E%7B%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%5Cln%20n%7D%7Bn%7D%7D$

$L=e^{0}=1$

那么,原式极限结果为1

$https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B1+%5Csqrt%7B2%7D+%5Csqrt%7B3%7D+...+%5Csqrt%7Bn%7D%7D%7B%5Csqrt%7Bn+1%7D+%5Csqrt%7Bn+2%7D+...%5Csqrt%7Bn+n%7D%7D$ （ $https://latex.csdn.net/eq?%5Cfrac%7B%5Cinfty%7D%7B%5Cinfty%7D$ ）

解: 设 $a_n=1+\sqrt{2}+\sqrt{3}+…+\sqrt{n}=\sum_{k=1}^{n}\sqrt{k}$

$a_{n-1}=\sum_{k=1}^{n-1}\sqrt{k}$

$b_n=\sqrt{n+1}+\sqrt{n+2}+…\sqrt{n+n}$

( 每一项内第一个n与其下标一致)

注意,对于来说,经过观察我们不难发现 $b_{n-1}$ 不单单意味着原数列的前n-1项,同时我们还应该将每一项内第一个n更改为n-1。即 $b_{n-1}=\sum_{k=1}^{n-1}\sqrt{n-1+k}$

$b_{n-1}=\sqrt{n-1+1}+\sqrt{n-1+2}+…+\sqrt{n-1+n-1}$

则 $a_n-a_{n-1}=\sqrt{n}$ ，

$b_n-b_{n-1}=\sqrt{2n}+\sqrt{2n-1}-\sqrt{n}$

$https://latex.csdn.net/eq?L%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Ba_%7Bn-1%7D%7D%7Bb_%7Bn-1%7D%7D%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%5Csqrt%7Bn%7D%7D%7B%5Csqrt%7B2n-1%7D+%5Csqrt%7B2n%7D-%5Csqrt%7Bn%7D%7D$

利用’抓大头’思想不难得到 $https://latex.csdn.net/eq?L%3D%5Cfrac%7B1%7D%7B2%5Csqrt%7B2%7D-1%7D$

那么,原式极限结果= $https://latex.csdn.net/eq?%5Cfrac%7B1%7D%7B2%5Csqrt%7B2%7D-1%7D$

$https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto%5Cinfty%7Dn%5Cbegin%7Bpmatrix%7D%20%5Csum_%7Bk%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bn+k%7D-ln2%20%5Cend%7Bpmatrix%7D$ （ $https://latex.csdn.net/eq?%5Cfrac%7B0%7D%7B0%7D$ ）

原式= $https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%5Cbegin%7Bpmatrix%7D%20%5Csum_%7Bk%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bn+k%7D-ln2%20%5Cend%7Bpmatrix%7D%7D%7B%5Cfrac%7B1%7D%7Bn%7D%7D$

设 $https://latex.csdn.net/eq?a_n%3D%5Csum_%7Bk%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bn+k%7D$ , $https://latex.csdn.net/eq?b_n%3D%5Cfrac%7B1%7D%7Bn%7D$

对于分子来说 $https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto%5Cinfty%7D%5Csum%20_%7Bk%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bn+k%7D$
可以变形为 $https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bk%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7B1+%5Cfrac%7Bk%7D%7Bn%7D%7D%3D%5Cint_%7B0%7D%5E%7B1%7D%5Cfrac%7B1%7D%7B1+x%7Ddx%3Dln2$
对于分母来说 $https://latex.csdn.net/eq?%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B1%7D%7Bn%7D%3D0$
因此该极限满足 $https://latex.csdn.net/eq?%5Cfrac%7B0%7D%7B0%7D$ 型未定式

令 $https://latex.csdn.net/eq?L%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Ba_%7Bn-1%7D%7D%7Bb_%7Bn-1%7D%7D$

$https://latex.csdn.net/eq?L%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%5Csum_%7Bk%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bn+k%7D-ln2-%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7D%5Cfrac%7B1%7D%7Bn-1+k%7D+ln2%7D%7B%5Cfrac%7B1%7D%7Bn%7D-%5Cfrac%7B1%7D%7Bn-1%7D%7D$

$https://latex.csdn.net/eq?L%3Dlim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%5Cfrac%7B1%7D%7B2n%7D+%5Cfrac%7B1%7D%7B2n-1%7D-%5Cfrac%7B1%7D%7Bn%7D%7B%7D%7D%7B%5Cfrac%7B-1%7D%7Bn%28n-1%29%7D%7D$