试证：lim(n→∞){n／[1／2+2／3+…+n／(n+1)]}^n／n=e^(γ-1)

elim

题：试证\(\;\displaystyle\lim_{n\to\infty}{\small\frac{1}{n}\big(\frac{n}{\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}}\big)^n=\;}e^{\gamma-1}\)

elim

题：试证\(\;\displaystyle\lim_{n\to\infty}{\small\frac{1}{n}\big(\frac{n}{\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}}\big)^n=\;}e^{\gamma-1}\)
证：\(\because\underset{\,}{\;}\frac{1}{2}+\cdots+\frac{n}{n+1}\small=n+1-H_{n+1}=n-(\gamma_{n+1}+\ln(n+1)-1)\)
\(\therefore\underset{\,}{\quad}x_n=\frac{1}{n}\big(\frac{1}{1-a_n}\big)^n,\;\;(na_n{\small=\gamma_{n+1}-1+\ln(n+1)=O(\ln n)}).\)
\(\therefore\underset{\,}{\quad}\ln x_n=n\ln(1-a_n)-\ln n=n(a_n+O(a_n^2))-\ln n\)
\(\qquad=\gamma_{n+1}-1+\ln(1+\frac{1}{n})+O(\frac{\ln^2 n}{n})\to\gamma-1.\quad\small\square\)

【LSFI】(1.2)

elim

awei 喜欢带\(\,\gamma\,\)的问题.

王守恩

elim 发表于 2020-11-13 12:49
awei 喜欢带\(\,\gamma\,\)的问题.

elim！能把如此重要的两个数学常数绞在一起，太好了！

\(\displaystyle\lim_{n\to\infty}\ \frac{1}{n}\bigg(\frac{n}{\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n} }\bigg)^n=e^{\ \gamma} \)

luyuanhong

楼上 elim 的帖子很好！已收藏。

elim

题：试证\(\;\;\displaystyle\lim_{n\to\infty}{\small\frac{\scriptsize 1}{n}\big(\frac{n}{\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-k-1}{n-k}}\big)^n=\;}e^{\gamma+k}\;\;\small(k\in\mathbb{Z}).\)
证：\(\because\underset{\,}{\;}\frac{1}{2}+\cdots+\frac{n-k-1}{n-k}\small=n-k-H_{n-k-1}\)
\(\qquad\qquad\qquad\qquad\qquad\;\small=n-(\gamma_{n-k-1}+\ln(n-k-1)+k)\)
\(\therefore\underset{\,}{\quad}x_n=\frac{1}{n}\big(\frac{1}{1-a_n}\big)^n_,\;(na_n{\small=\gamma_{n-k-1}+k+\ln(n-k-1)=O(\ln n)}).\)
\(\therefore\underset{\,}{\quad}\ln x_n=-n\ln(1-a_n)-\ln n=n(a_n+O(a_n^2))-\ln n\)
\(\qquad=\gamma_{n+1}+k+\ln(1-\frac{k+1}{n})+O(\frac{\ln^2 n}{n})\to\gamma+k.\quad\small\square\)

注记：\(\displaystyle\lim_{n\to\infty}{\small\frac{\scriptsize 1}{n}\big(\frac{n}{\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n}}\big)^n=\;}e^{\gamma}\)

luyuanhong

楼上 elim 的帖子很好！已收藏。

王守恩

elim 发表于 2020-11-14 03:19
题：试证\(\;\;\displaystyle\lim_{n\to\infty}{\small\frac{\scriptsize 1}{n}\big(\frac{n}{\frac{1}{2}+ ...

\(\displaystyle\lim_{n\to\infty}\frac{1}{n}\big(\frac{n}{\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n}}\big)^n=e^{\gamma}\)
\(\displaystyle\lim_{n\to\infty}\frac{1}{n}\big(\frac{n}{\frac{2}{3}+\frac{3}{4}+\cdots+\frac{n-1}{n}}\big)^n=e^{\gamma+\frac{1}{2}}\)
\(\displaystyle\lim_{n\to\infty}\frac{1}{n}\big(\frac{n}{\frac{3}{4}+\frac{4}{5}+\cdots+\frac{n-1}{n}}\big)^n=e^{\gamma+\frac{7}{6}}\)
\(\displaystyle\lim_{n\to\infty}\frac{1}{n}\big(\frac{n}{\frac{4}{5}+\frac{5}{6}+\cdots+\frac{n-1}{n}}\big)^n=e^{\gamma+\frac{23}{12}}\)
\(\displaystyle\lim_{n\to\infty}\frac{1}{n}\big(\frac{n}{\frac{5}{6}+\frac{6}{7}+\cdots+\frac{n-1}{n}}\big)^n=e^{\gamma+\frac{163}{60}}\)
\(\displaystyle\lim_{n\to\infty}\frac{1}{n}\big(\frac{n}{\frac{6}{7}+\frac{7}{8}+\cdots+\frac{n-1}{n}}\big)^n=e^{\gamma+\frac{71}{20}}\)
\(\displaystyle\lim_{n\to\infty}\frac{1}{n}\big(\frac{n}{\frac{7}{8}+\frac{8}{9}+\cdots+\frac{n-1}{n}}\big)^n=e^{\gamma+\frac{617}{140}}\)

王守恩

王守恩 发表于 2020-11-14 07:38
\(\displaystyle\lim_{n\to\infty}\frac{1}{n}\big(\frac{n}{\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1} ...

elim！能把如此重要的两个数学常数绞在一起，太不容易了！

\(\displaystyle\lim_{n\to\infty}\bigg(\sum_{k=1}^{n}\ \frac{\sqrt[n]{n}(k-1)}{n\ k}\bigg)^{n}=\frac{1}{e^{\gamma}}\)

wangyangke

王守恩老师在论坛是明星闪耀了！王守恩老师就是曾经的awi或gbn网友吧，，，