悠闲数学娱乐论坛(第2版)'s Archiver

青青子衿 发表于 2019-2-12 17:39

渐近展开组合数对数形成的求和式

[i=s] 本帖最后由 青青子衿 于 2019-2-13 16:57 编辑 [/i]

\[ \sum\limits_{k=0}^n\ln\binom{n}{k}=\frac{n\left(n+2\right)}{2}-\frac{\left(3n+2\right)\ln\left(2\pi n\right)}{6}-\frac{1}{12}+\frac{\gamma}{6}+\frac{1}{\pi^2}\sum\limits_{k=0}^\infty\frac{\ln k}{k^2}+o\left(\frac{1}{n}\right) \]

战巡 发表于 2019-2-13 08:04

[i=s] 本帖最后由 战巡 于 2019-2-13 10:12 编辑 [/i]

[b]回复 [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=29824&ptid=5890]1#[/url] [i]青青子衿[/i] [/b]

首先我们有
\[\frac{1}{\pi^2}\sum_{k=1}^\infty\frac{\ln(k)}{k^2}=2\ln(A)-\frac{\gamma}{6}-\frac{\ln(2\pi)}{6}\]
$A$为格莱舍常数,这个在[url]http://kuing.orzweb.net/viewthread.php?tid=3980[/url]已经证明过

而后
\[\frac{n(n+2)}{2}-\frac{(3n+2)\ln(2\pi n)}{6}-\frac{1}{12}+\frac{\gamma}{6}+\frac{1}{\pi^2}\sum_{k=1}^\infty\frac{\ln(k)}{k^2}\]
\[=\frac{n(n+2)}{2}-\frac{(3n+2)\ln(n)}{6}-\frac{1}{12}+2\ln(A)-\frac{(n+1)\ln(2\pi)}{2}\]

左边有
\[\sum_{k=0}^n\ln(C^k_n)=\sum_{k=0}^n\ln(\frac{n!}{k!(n-k)!})=\ln(\frac{(n!)^{n+1}}{(1!·2!·...·n!)^2})\]
\[=\ln(\frac{(n!)^{n-1}}{(1!·2!·...·(n-1)!)^2})=\ln(\frac{\Gamma(n+1)^{n-1}}{G(n+1)^2})=\ln(\frac{K(n+1)^2}{\Gamma(n+1)^{n+1}})\]
\[=2\ln(K(n+1))-(n+1)\ln(\Gamma(n+1))\]
\[=[2\ln(A)-\frac{n^2}{2}+(n(n+1)+\frac{1}{6})\ln(n)+2\ln(1+\frac{1}{720n^2}+...)]\]
\[-[(n+1)^2\ln(n+1)-(n+1)^2-\frac{n+1}{2}\ln(n+1)+\frac{n+1}{2}\ln(2\pi)+\frac{1}{12}+o(\frac{1}{n})]\]
\[=2\ln(A)-\frac{3n+2}{6}\ln(n)-\frac{1}{12}-\frac{n+1}{2}\ln(2\pi)+\frac{n^2}{2}+2n+1-\frac{(n+1)(2n+1)}{2}\ln(1+\frac{1}{n})+o(\frac{1}{n})\]
\[=2\ln(A)-\frac{3n+2}{6}\ln(n)-\frac{1}{12}-\frac{n+1}{2}\ln(2\pi)+\frac{n^2}{2}+2n+1-\frac{(n+1)(2n+1)}{2}(\frac{1}{n}-\frac{1}{2n^2}+o(\frac{1}{n^2}))+o(\frac{1}{n})\]
\[=2\ln(A)-\frac{3n+2}{6}\ln(n)-\frac{1}{12}-\frac{n+1}{2}\ln(2\pi)+\frac{n(n+2)}{2}+o(\frac{1}{n})\]

青青子衿 发表于 2019-2-13 09:45

[i=s] 本帖最后由 青青子衿 于 2019-2-13 09:49 编辑 [/i]

[quote]回复  青青子衿
首先我们有
\[\frac{1}{\pi^2}\sum_{k=1}^\infty\frac{\ln(k)}{k^2}=2\ln(A)-\frac{\gamma}{6}-\frac{\ln(2\pi)}{6}\] ...
[size=2][color=#999999]战巡 发表于 2019-2-13 08:04[/color] [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=29836&ptid=5890][img]http://kuing.orzweb.net/images/common/back.gif[/img][/url][/size][/quote]
谢谢站版,可是这两个有一项不一样,不影响吗?
这个渐近展开的结果是从一篇文章上看到的的,也没有给出证明……
\begin{align*}
&\frac{n(n+2)}{2}-\frac{(3n+2)\ln(2\pi n)}{6}-\frac{1}{12}+\frac{\gamma}{6}+\frac{1}{\pi^2}\sum_{k=1}^\infty\frac{\ln(k)}{k^2}\\
=\,&\frac{n(n+2)}{2}-\frac{(3n+2)\ln(n)}{6}-\frac{(3n+2)\ln(2\pi)}{6}-\frac{1}{12}+\frac{\gamma}{6}+\frac{1}{\pi^2}\sum_{k=1}^\infty\frac{\ln(k)}{k^2}\\
=\,&\frac{n(n+2)}{2}-\frac{(3n+2)\ln(n)}{6}-\frac{(3n+3)\ln(2\pi)}{6}-\frac{1}{12}+\frac{\ln(2\pi)}{6}+\frac{\gamma}{6}+\frac{1}{\pi^2}\sum_{k=1}^\infty\frac{\ln(k)}{k^2}\\
=\,&\frac{n(n+2)}{2}-\frac{(3n+2)\ln(n)}{6}-\frac{(n+1)\ln(2\pi)}{2}-\frac{1}{12}+2\ln(A)
\end{align*}

\begin{align*}
&\sum\limits_{k=0}^n\ln\binom{n}{k}\\
=\,&\cdots\\
=\,&\cdots\\
=\,&\frac{n(n+2)}{2}-\frac{(3n+2)\ln(n)}{6}-\frac{1}{12}+2\ln(A)+o\left(\frac{1}{n}\right)
\end{align*}

战巡 发表于 2019-2-13 10:14

[b]回复 [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=29838&ptid=5890]3#[/url] [i]青青子衿[/i] [/b]


哦....漏写了而已...
前面过程中有的,后面合并项逐段处理的时候漏了...

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