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
$$\int_{0}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}=2\ln{\pi}-\ln{4}$$

$\square$