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Last edited by hbghlyj at 2023-8-6 18:40:00Gamma.zip中的Mascheroni.txt定义了
\begin{align*}L(n)&= \sum_{k=1}^n \frac1k - \ln(n) - \frac1{2n}\\R(n)&=\sum_{k=1}^n \frac1k - \ln(n)- \frac1{2(n+1)}\end{align*}
容易看出$$\lim_{{n \to \infty}} L(n)= \lim_{{n \to \infty}} R(n)= \lim_{{n \to \infty}} \sum_{k=1}^{n} \frac{1}{k} - \ln(n) = \gamma$$
下面要证明 \(L(n)\) 单增, \(R(n)\) 单减.
首先对$n+1$和$n$使用调和平均$\le$对数平均$\le$算术平均
\begin{equation}\label1
\frac1{n + \frac12}<\ln(n + 1) -\ln(n) < \frac{\frac1n + \frac1{n+1}} 2
\end{equation}
为证明 \(L(n)\) 单增,
\begin{align*}
L(n+1)&= \sum_{k=1}^{n+1} \frac{1}{k} - \ln(n+1) - \frac{1}{2(n+1)}\\
&= \sum_{k=1}^{n} \frac{1}{k} - \ln(n) - \frac{1}{2n} + \frac{1}{n+1} - \left[\ln(n+1) - \ln(n)\right] + \frac{1}{2n} - \frac{1}{2(n+1)}
\end{align*}
由\eqref{1}右边
\begin{align*}
L(n+1) &> L(n) + \frac{1}{n+1} - \frac{1}{2n} - \frac{1}{2(n+1)} + \frac{1}{2n} - \frac{1}{2(n+1)}
\\& = L(n)
\end{align*}
为证明 \(R(n)\) 单减,
\begin{align*}
R(n+1)&= \sum_{k=1}^{n+1} \frac{1}{k} - \ln(n+1) - \frac{1}{2(n+2)}\\
&= \sum_{k=1}^{n} \frac{1}{k} - \ln(n) - \frac{1}{2(n+1)} + \frac{1}{n+1} - \left[\ln(n+1) - \ln(n)\right] + \frac{1}{2(n+1)} - \frac{1}{2(n+2)}
\end{align*}
由\eqref{1}左边
\begin{align*}
R(n+1)&< R(n) + \frac{1}{n+1} - \frac{1}{\frac{n+1}{2}} + \frac{1}{2(n+1)} - \frac{1}{2(n+2)}
\\&= R(n) - \frac{3}{2(n+1)(n+2)(n+1/2)}\\&< R(n)
\end{align*} |
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kuing
Posted at 2023-8-6 18:14:32
