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令$\,H_n = {\small\displaystyle\sum_{k=1}^n}\large\frac{1}{k}\;$(调和级数的前$\,n\,$项部分和).则$\,H_{2^{n+1}}-H_{2^n}\ge\large\frac{1}{2},$&h
$\quad H_{2^n}\ge 1+{\large\frac{n}{2}},\;{\displaystyle\lim_{n\to\infty}}H_n = \infty$ (调和级数发散的初等证明). 由JG
$\displaystyle{\quad 0< H_n-1<{\small\int_1^n\frac{dx}{\lfloor x\rfloor +1}}}< \ln n={\small\int_1^n\frac{dx}{x}}< {\small\int_1^n\frac{dx}{\lfloor x\rfloor}}=H_{n-1}\;$得{F9x+
$\quad 0< H_{n-1}-\ln n < \overset{\,}{H_n} -\ln n < 1,\;$（远比${\displaystyle{\,\lim_{n\to\infty}}}{\large\frac{\overset{\,}{H_n}}{\ln n}}=1\,$精致.）?).L
$\underset{\,}{\because}\,\Delta(H_n-\ln n)=\frac{1}{n+1}(1-\ln(1+\frac{1}{n})^{n+1})< 0\;\small(\{(1+\frac{1}{n})^{n+1}\}\,\text{严格减}).$;%IQA
$\underset{\,}{\therefore}\;\exists\,\gamma={\displaystyle\lim_{n\to\infty}}(H_n-\ln n)=0.5772156649015328606065120900824\ldots$*Ovxu
$\underset{\,}{\quad}$对$\,\Delta H_n =\large\frac{1}{n}-\frac{1}{n^2}+\cdots+\frac{\small(-1)^{k-1}}{n^k}+{\small O}(\frac{1}{n^{n+1}})\;$两边取$\;\Delta^{-1}\,$得&:
$\quad\boxed{H_n = \ln n +\gamma +\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4}-\frac{1}{252n^6}+O(\frac{1}{n^8})}$W}j
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2018/04/17 09:05am 
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