证明
\[\lim_{n\rightarrow\infty}{\frac{\ln{n}}{n}\left(\frac{\sum_{k=1}^{n=1}{\csc{\left(\frac{k\pi}{n}\right)}}}{\ln{n}}-\frac{2}{\pi}n\right)}=\frac{2\gamma}{\pi}-\frac{2\ln{\pi}-\ln{4}}{\pi}\]

(tian_275461)

证明
记
\begin{align*}
 I&= \frac{\ln{n}}{n}\left(\frac{\sum_{k=1}^{n-1}{\csc{\left(\frac{k\pi}{n}\right)}}}{\ln{n}}-\frac{2}{\pi}n\right)\\
 &=\frac{\pi}{n}\sum_{k=1}^{n-1}\csc{\left(\frac{k\pi}{n}\right)}-2\ln{n}
\end{align*}
只要证
\[ \lim_{n\rightarrow\infty}{I}=2\gamma-2\ln{\pi}+\ln{4} \]
也就是
\[ \lim_{n\rightarrow\infty}{(2\gamma-I)}=2\ln{\pi}-\ln{4} \]
记$ S=2\gamma-I$,我们有
\[ \gamma=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}-\ln{n}+c_{n} \quad \text{其中} c_{n}\rightarrow 0\quad  (n\rightarrow\infty) \]
\begin{align*}
 S&=2\sum_{k=1}^{n-1}-\frac{\pi}{n}\sum_{k=1}^{n-1}{\csc{\left(\frac{k\pi}{n}\right)}}+2c_{n}\\
 &=\sum_{k=1}^{n-1}{\left(\frac{1}{k}+\frac{1}{n-k}\right)}-\frac{\pi}{n}\sum_{k=1}^{n-1}{\csc{\left(\frac{k\pi}{n}\right)}}+2c_{n}\\
 &=\frac{\pi}{n}\sum_{k=1}^{n-1}{\left(\dfrac{1}{\frac{k\pi}{n}}+\dfrac{1}{\pi-\frac{k\pi}{n}}\right)}-\frac{\pi}{n}\sum_{k=1}^{n-1}{\csc{\left(\frac{k\pi}{n}\right)}}+2c_{n}
\end{align*}
所以有
\[  \lim_{n\rightarrow\infty}{S}=\int_{0}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx} \]
故只要证
\[ \int_{0}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}=2\ln{\pi}-\ln{4}\]
而
\[ \int_{0}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}=\int_{0}^{\frac{\pi}{2}}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}+\int_{\frac{\pi}{2}}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}\]
 第二部分用替换$ y=\pi-x $
\[ \Rightarrow \lim_{n\rightarrow\infty}{S}=2\int_{0}^{\frac{\pi}{2}}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}\]
注意到
\[ \frac{1}{\sin{x}}=\frac{1}{x}+\sum_{n=1}^{\infty}{(-1)^{n}\left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right)} \]
\begin{align*}
 \int_{0}^{\frac{\pi}{2}}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}&=\sum_{n=1}^{\infty}{(-1)^{n+1}\int_{0}^{\frac{\pi}{2}}{\left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}   \right)dx}}\\
 &=\sum_{n=1}^{\infty}{(-1)^{n+1}\cdot\ln{\left(1-\frac{1}{4n^2}\right)}}
\end{align*}
对$n$分奇偶性讨论
(1) $n=2m-1\ (m=1,2,\cdots) $ 时
\[ (-1)^{n+1}\cdot\ln{\left(1-\frac{1}{4n^2}\right)}=\ln{\left(\frac{(4m-1)(4m-3)}{(4m-2)^2}\right)}\]
(2) $n=2m\ (m=1,2,\cdots) $ 时
\[ (-1)^{n+1}\cdot\ln{\left(1-\frac{1}{4n^2}\right)}=\ln{\left(\frac{(4m)^2}{(4m+1)(4m-1)}\right)}\]
而
\[ \sum_{m=1}^{k}{(-1)^{n+1}\cdot\ln{\left(1-\frac{1}{4n^2}\right)}}=\ln{\left[\frac{1}{4k+1}\left(\frac{(2k)!!}{(2k-1)!!}\right)^{2}\right]}\rightarrow \ln{\frac{\pi}{4}} \quad \text{(用Wallis公式)} \]
马上得到
\[ \int_{0}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}=2\ln{\pi}-\ln{4}\]

$\square$