【數(shù)學(xué)分析】一道簡(jiǎn)單數(shù)列極限的多種解法

2023-04-11 23:00 作者:Ice_koucha 0人讀過(guò) | 我要投稿

前言：今天突然心血來(lái)潮打算開(kāi)這個(gè)坑，大概是記錄自己一些想法和當(dāng)做復(fù)習(xí)筆記的作用，之前已經(jīng)做了兩個(gè)數(shù)學(xué)的視頻，但是做視頻有點(diǎn)累，而且不蹭熱點(diǎn)就沒(méi)什么再生數(shù)。所以這次嘗試用專欄的形式發(fā)出來(lái)，隨便使用一下B站專欄的公式編輯功能，更新完全隨緣，既然是第一題，那么就選一個(gè)非常簡(jiǎn)單的問(wèn)題來(lái)講解，(會(huì)用四種不同的方法來(lái)解答)話不多說(shuō)，我們開(kāi)始吧(為獲得最佳閱讀體驗(yàn)，建議在PC網(wǎng)頁(yè)端瀏覽)。

題目：設(shè) $a_%7Bn%7D%3D%5Cfrac%7B1%7D%7Bn%2B1%7D%2B%5Cfrac%7B1%7D%7Bn%2B2%7D%2B%5Ccdots%2B%5Cfrac%7B1%7D%7Bn%2Bn%7D$ ，求 $%20%5Clim_%7Bn%5Cto%E2%88%9E%7D%20%20a_%7Bn%7D%20$

分析：為了嚴(yán)謹(jǐn)起見(jiàn)，我們先證明 $a_%7Bn%7D%20$ 收斂.

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 $a_%7Bn%7D%20$ 單調(diào)遞增.

又因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=a_%7Bn%7D%20%3C%5Cfrac%7Bn%7D%7Bn%2B1%7D%20%20%3C1" alt="a_%7Bn%7D%20%3C%5Cfrac%7Bn%7D%7Bn%2B1%7D%20%20%3C1">，所以 $a_%7Bn%7D%20$ 單調(diào)遞增且有上界，由單調(diào)有界原則， $a_%7Bn%7D%20$ 收斂.

解法一(利用 $Riemann$ 積分的定義)

最簡(jiǎn)單直接的做法，沒(méi)什么好說(shuō)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 $Euler$ 常數(shù))

先給出下列命題：

①對(duì)數(shù)不等式： $%5Cfrac%7Bx%7D%7B1%2Bx%7D%5Cleq%20%20ln(x%2B1)%5Cleq%20x$ ?(當(dāng) $x%3E-1$ 時(shí))

證明：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 $0$ 與 $x$ 之間，不等式右端得證.對(duì)于左端使用類似的方法也可證出.當(dāng)且僅當(dāng) $x%3D0$ 時(shí)兩邊取等號(hào).

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 $n$ 為正整數(shù)

②數(shù)列 $b_%7Bn%7D%3D1%2B%5Cfrac%7B1%7D%7B2%7D%20%2B%5Cfrac%7B1%7D%7B3%7D%20%2B%5Ccdots%2B%5Cfrac%7B1%7D%7Bn%7D%20-lnn$ 收斂

證明：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 $b_%7Bn%7D%20$ 單調(diào)遞減。

同時(shí)，

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

所以 $b_%7Bn%7D%20$ 單調(diào)遞減且有下界，由單調(diào)有界原則，知 $b_%7Bn%7D%20$ 收斂.我們把 $b_%7Bn%7D%20$ 的極限記作 $%5Cgamma%20$ ，稱為 $Euler$ 常數(shù).

因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=b_%7B2n%7D%20" alt="b_%7B2n%7D%20">是 $b_%7Bn%7D%20$ 的子列，所以ta們會(huì)收斂到相同的極限.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(利用夾逼定理)

由解法二的①知，有

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

化簡(jiǎn)，得

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

又因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%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" alt="%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">

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

由夾逼定理知

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

解法四(利用 $Catalan$ 恒等式)

④( $Catalan$ 恒等式)

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

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

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

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

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

令 $2n%5Crightarrow%20%E2%88%9E$ ，注意到 $ln(x%2B1)$ 的 $Maclaurin$ 級(jí)數(shù)展開(kāi)為

$ln(x%2B1)%3D%5Csum_%7Bn%3D0%7D%5E%E2%88%9E%5Cfrac%7B(-1)%5En%20%7D%7Bn%2B1%7D%20%20x%5E%7Bn%2B1%7D$ ，收斂域?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=(-1%2C1%5D" alt="(-1%2C1%5D">，取 $x%3D1$ ，有

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