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战巡

求极限：
\[\lim_{n\to+\infty}\frac{\ln(n)}{n}[\frac{\sum_{k=1}^{n-1}\csc(\frac{k\pi}{n})}{\ln(n)}-\frac{2n}{\pi}]\]
\[\lim_{n\to+\infty}\frac{\ln(n)}{n}[\frac{\sum_{k=1}^{n-1}\csc(\frac{k\pi}{n})}{\ln(n)}-\frac{2n}{\pi}]=\lim_{n\to+\infty}[\frac{\sum_{k=1}^{n-1}\csc(\frac{k\pi}{n})}{n}-\frac{2\ln(n)}{\pi}]\]
\[=\lim_{n\to+\infty}[\frac{\sum_{k=1}^{n-1}(\csc(\frac{k\pi}{n})-\frac{2n}{k\pi})}{n}+\sum_{k=1}^{n-1}\frac{2}{k\pi}-\frac{2\ln(n)}{\pi}]\]
\[=\lim_{n\to+\infty}[\frac{\sum_{k=1}^{n-1}(\csc(\frac{k\pi}{n})-\frac{2n}{k\pi})}{n}]+\frac{2\gamma}{\pi}=\frac{1}{\pi}\int_{0}^{\pi}(\csc(x)-\frac{2}{x})dx+\frac{2\gamma}{\pi}\]
\[=\frac{1}{\pi}[\lim_{a\to \pi}\ln(\frac{\tan(\frac{a}{2})}{a^2})-\lim_{b\to 0}\ln(\frac{\tan(\frac{b}{2})}{b^2})]+\frac{2\gamma}{\pi}\]
\[=\frac{1}{\pi}[\lim_{a\to 0}\ln(\frac{\tan(\frac{\pi-a}{2})}{(\pi-a)^2}\frac{a^2}{\tan(\frac{a}{2})})]+\frac{2\gamma}{\pi}=\frac{\ln(\frac{4}{\pi^2})}{\pi}+\frac{2\gamma}{\pi}\]

青青子衿

回复 1# 战巡
Mark一下
pxchg1200.is-programmer.com/posts/39456.html

hbghlyj

Computation of limit in the last line:
\[\frac{\tan(\frac{\pi-a}{2})}{(\pi-a)^2}\cdot\frac{a^2}{\tan(\frac{a}{2})}=\frac{1}{(\pi-a)^2}\cdot\frac{a^2}{\tan(\frac{a}{2})^2}\to\frac1{\pi^2}\cdot4\]