【数学分析】一道简单数列极限的多种解法

2023-04-11 23:00 作者:Ice_koucha 0人读过 | 我要投稿

前言：今天突然心血来潮打算开这个坑，大概是记录自己一些想法和当做复习笔记的作用，之前已经做了两个数学的视频，但是做视频有点累，而且不蹭热点就没什么再生数。所以这次尝试用专栏的形式发出来，随便使用一下B站专栏的公式编辑功能，更新完全随缘，既然是第一题，那么就选一个非常简单的问题来讲解，(会用四种不同的方法来解答)话不多说，我们开始吧(为获得最佳阅读体验，建议在PC网页端浏览)。

题目：设 $a_%7Bn%7D%3D%5Cfrac%7B1%7D%7Bn%2B1%7D%2B%5Cfrac%7B1%7D%7Bn%2B2%7D%2B%5Ccdots%2B%5Cfrac%7B1%7D%7Bn%2Bn%7D$ ，求 $%20%5Clim_%7Bn%5Cto%E2%88%9E%7D%20%20a_%7Bn%7D%20$

分析：为了严谨起见，我们先证明 $a_%7Bn%7D%20$ 收敛.

先作差，有 $a_%7Bn%2B1%7D%20-a_%7Bn%7D%3D(%5Cfrac%7B1%7D%7Bn%2B2%7D%20%2B%5Cfrac%7B1%7D%7Bn%2B3%7D%2B%5Ccdots%2B%5Cfrac%7B1%7D%7B2n%2B2%7D)-(%5Cfrac%7B1%7D%7Bn%2B1%7D%2B%5Cfrac%7B1%7D%7Bn%2B2%7D%2B%5Ccdots%2B%5Cfrac%7B1%7D%7Bn%2Bn%7D)%0A$

$%3D%5Cfrac%7B1%7D%7B2n%2B1%7D-%20%5Cfrac%7B1%7D%7B2n%2B2%7D%3E0$

因此 $a_%7Bn%7D%20$ 单调递增.

又因为 $a_%7Bn%7D%20%3C%5Cfrac%7Bn%7D%7Bn%2B1%7D%20%20%3C1$ ，所以 $a_%7Bn%7D%20$ 单调递增且有上界，由单调有界原则， $a_%7Bn%7D%20$ 收敛.

解法一(利用 $Riemann$ 积分的定义)

最简单直接的做法，没什么好说的

$%5Clim_%7Bn%5Cto%E2%88%9E%7D%20a_%7Bn%7D%20%20%3D%5Clim_%7Bn%5Cto%E2%88%9E%7D%20%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%5Cfrac%7B1%7D%7B1%2B%5Cfrac%7Bi%7D%7Bn%7D%20%7D%20%7D%7Bn%7D%20%3D%5Cint_%7B0%7D%5E%7B1%7D%20%5Cfrac%7B1%7D%7Bx%2B1%7Ddx%3Dln2%20$

解法二(利用 $Euler$ 常数)

先给出下列命题：

①对数不等式： $%5Cfrac%7Bx%7D%7B1%2Bx%7D%5Cleq%20%20ln(x%2B1)%5Cleq%20x$ (当 $x%3E-1$ 时)

证明：令 $f(x)%3Dx-ln(x%2B1)$ ，注意到 $f(0)%3D0$ ，由 $Lagrange$ 中值定理得，

$f(x)%3D%5Cfrac%7B%5Cxi%20x%7D%7B%5Cxi%2B1%7D%20%3E0$ ，其中 $%5Cxi%20$ 介于 $0$ 与 $x$ 之间，不等式右端得证.对于左端使用类似的方法也可证出.当且仅当 $x%3D0$ 时两边取等号.

取 $x%3D%5Cfrac%7B1%7D%7Bn%7D%20$ ，得 $%5Cfrac%7B1%7D%7Bn%2B1%7D%20%3Cln(1%2B%5Cfrac%7B1%7D%7Bn%7D)%3C%5Cfrac%7B1%7D%7Bn%7D%20$ ，其中 $n$ 为正整数

②数列 $b_%7Bn%7D%3D1%2B%5Cfrac%7B1%7D%7B2%7D%20%2B%5Cfrac%7B1%7D%7B3%7D%20%2B%5Ccdots%2B%5Cfrac%7B1%7D%7Bn%7D%20-lnn$ 收敛

证明：作差，得

$b_%7Bn%2B1%7D-b_%7Bn%7D%20%20%3D%5Cfrac%7B1%7D%7Bn%2B1%7D%20-ln(n%2B1)-lnn$

$%3D%5Cfrac%7B1%7D%7Bn%2B1%7D-ln%20(1%2B%5Cfrac%7B1%7D%7Bn%7D%20)$

由①得， $%5Cfrac%7B1%7D%7Bn%2B1%7D-ln%20(1%2B%5Cfrac%7B1%7D%7Bn%7D%20)%3C0$ ，所以 $b_%7Bn%7D%20$ 单调递减。

同时，

$b_%7Bn%7D%20%3Eln(1%2B1)%2Bln(1%2B%5Cfrac%7B1%7D%7B2%7D%20)%2Bln(1%2B%5Cfrac%7B1%7D%7B3%7D%20)%2B%5Ccdots%2Bln(1%2B%5Cfrac%7B1%7D%7Bn%7D%20)-lnn%3Dln(1%2B%5Cfrac%7B1%7D%7Bn%7D%20)%3E0$

所以 $b_%7Bn%7D%20$ 单调递减且有下界，由单调有界原则，知 $b_%7Bn%7D%20$ 收敛.我们把 $b_%7Bn%7D%20$ 的极限记作 $%5Cgamma%20$ ，称为 $Euler$ 常数.

因为 $b_%7B2n%7D%20$ 是 $b_%7Bn%7D%20$ 的子列，所以ta们会收敛到相同的极限.于是有：

$%5Clim_%7Bn%5Cto%E2%88%9E%7D%20a_%7Bn%7D%20%20%3D%5Clim_%7Bn%5Cto%E2%88%9E%7D%20(b_%7B2n%7D%2Bln2n)-(b_%7Bn%7D%2Blnn)$

$%3D%5Clim_%7Bn%5Cto%E2%88%9E%7D%20(%5Cgamma%20-%5Cgamma)%2B%20(lnn-lnn)%2Bln2$

$%3Dln2$

解法三(利用夹逼定理)

由解法二的①知，有

$ln(1%2B%5Cfrac%7B1%7D%7Bn%2B1%7D%20)%2Bln(1%2B%5Cfrac%7B1%7D%7Bn%2B2%7D%20)%2B%5Ccdots%2Bln(1%2B%5Cfrac%7B1%7D%7B2n%7D%20)%3Ca_%7Bn%7D%20%3Cln(1%2B%5Cfrac%7B1%7D%7Bn%7D%20)%2Bln(1%2B%5Cfrac%7B1%7D%7Bn%2B1%7D%20)%2B%5Ccdots%2Bln(1%2B%5Cfrac%7B1%7D%7B2n-1%7D%20)$

化简，得

$ln(2n%2B1)-ln(n%2B1)%3Ca_%7Bn%7D%20%3Cln2n-lnn$

又因为 $%5Clim_%7Bn%5Cto%E2%88%9E%7D%20ln(2n%2B1)-ln(n%2B1)%3D%5Clim_%7Bn%5Cto%E2%88%9E%7Dln(%5Cfrac%7B2n%2B1%7D%7Bn%2B1%7D%20)%3Dln2$

$%5Clim_%7Bn%5Cto%E2%88%9E%7D%20ln2n-lnn%3D%5Clim_%7Bn%5Cto%E2%88%9E%7Dln2%2Blnn-lnn%3Dln2$

由夹逼定理知

$%5Clim_%7Bn%5Cto%E2%88%9E%7D%20a_%7Bn%7D%20%3Dln2$

解法四(利用 $Catalan$ 恒等式)

④( $Catalan$ 恒等式)

设 $c_%7B2n%7D%20%3D1-%5Cfrac%7B1%7D%7B2%7D%20%2B%5Cfrac%7B1%7D%7B3%7D-%5Cfrac%7B1%7D%7B4%7D%2B%5Ccdots%2B%5Cfrac%7B1%7D%7B2n-1%7D-%5Cfrac%7B1%7D%7B2n%7D$

则 $c_%7B2n%7D%20%3D(1%2B%5Cfrac%7B1%7D%7B2%7D%20%2B%5Cfrac%7B1%7D%7B3%7D%20%2B%5Ccdots%2B%5Cfrac%7B1%7D%7B2n-1%7D%2B%5Cfrac%7B1%7D%7B2n%7D)-2(%5Cfrac%7B1%7D%7B2%7D%20%2B%5Cfrac%7B1%7D%7B4%7D%2B%5Ccdots%2B%5Cfrac%7B1%7D%7B2n%7D)$

$%3D(1%2B%5Cfrac%7B1%7D%7B2%7D%20%2B%5Cfrac%7B1%7D%7B3%7D%20%2B%5Ccdots%2B%5Cfrac%7B1%7D%7B2n-1%7D%2B%5Cfrac%7B1%7D%7B2n%7D)-(1%2B%5Cfrac%7B1%7D%7B2%7D%2B%5Cfrac%7B1%7D%7B4%7D%2B%5Ccdots%2B%5Cfrac%7B1%7D%7Bn%7D%20)$

$%3D%5Cfrac%7B1%7D%7Bn%2B1%7D%2B%5Cfrac%7B1%7D%7Bn%2B2%7D%2B%5Ccdots%2B%5Cfrac%7B1%7D%7Bn%2Bn%7D$

$%3Da_%7Bn%7D%20$

令 $2n%5Crightarrow%20%E2%88%9E$ ，注意到 $ln(x%2B1)$ 的 $Maclaurin$ 级数展开为

$ln(x%2B1)%3D%5Csum_%7Bn%3D0%7D%5E%E2%88%9E%5Cfrac%7B(-1)%5En%20%7D%7Bn%2B1%7D%20%20x%5E%7Bn%2B1%7D$ ，收敛域为 $(-1%2C1%5D$ ，取 $x%3D1$ ，有

$ln(1%2B1)%3D1-%5Cfrac%7B1%7D%7B2%7D%20%2B%5Cfrac%7B1%7D%7B3%7D-%5Cfrac%7B1%7D%7B4%7D%2B%5Ccdots%2B%5Cfrac%7B1%7D%7B2n-1%7D-%5Cfrac%7B1%7D%7B2n%7D%3Dln2$