From bdd8fbd283fd2d24250ce383c4447e6c1f68ea74 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com>
Date: Mon, 26 Jan 2026 12:42:43 +0800
Subject: [PATCH] vault backup: 2026-01-26 12:42:43

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 笔记分享/自然对数的底与欧拉常数.md | 4 ++++
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+考虑两个数列$$x_n=\left(1+\frac{1}{n}\right)^n,y_n=\left(1+\frac{1}{n}\right)^{n+1},$$对所有的自然数 $n,$ 显然有 $\displaystyle y_n=x_n\left(1+\frac{1}{n}\right)>x_n.$ 接下来考察两个数列的单调性. 
+由“几何平均数小于等于算术平均数”可知$$x_n=\left(1+\frac{1}{n}\right)^n\cdot1\le\left(\frac{n(1+1/n)+1}{n+1}\right)^{n+1}=\left(1+\frac{1}{n+1}\right)^{n+1}=x_{n+1},$$于是数列 $\{x_n\}$ 单调递增; 
+$$\frac{1}{y_n}=\left(\frac{n}{n+1}\right)^{n+1}\cdot1\le\left(\frac{(n+1)\frac{n}{n+1}+1}{n+2}\right)^{n+2}=\left(\frac{n+1}{n+2}\right)^{n+2}=\frac{1}{y_{n+1}},$$于是数列 $\displaystyle\{\frac{1}{y_n}\}$ 单调递增，即数列 $\{y_n\}$ 单调递减. 于是有 $$2=x_1\le x_n<y_n\le y_1=4,$$故两个数列均有界. 由单调有界定理知两个数列均收敛. 又有 $y_n=x_n(1+\frac{1}{n}),$ 故两者极限相同，设为 $\text e$. 
+由上面的推导，我们有 $$x_n=\left(1+\frac{1}{n}\right)^n<\text e<\left(1+\frac{1}{n}\right)^{n+1}=y_n,$$两边取对数，整理得$$\frac{1}{n+1}<\ln\left(\frac{n+1}{n}\right)<\frac{1}{n},$$记 $\displaystyle b_n=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n,$ 则有$$b_{n+1}-b_n=\frac{1}{n+1}-\ln \frac{n+1}{n}<0,$$又有$$b_n>\ln \frac{2}{1}+\ln \frac{3}{2}+\cdots+\frac{n+1}{n}-\ln n=\ln (n+1)-\ln n>0,$$故数列 $\{b_n\}$ 单调递减有下界，故收敛，记 $\gamma=\lim\limits_{n\to\infty} b_n.$ 
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