\(H_n=\ln(n)+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4}-\frac{1}{252n^6}+\frac{1}{240n^8}-\ldots\)
我们可以得到：
\[\sum_{k=1}^n (H_k)^4=\sum_{k=1}^n (\ln(k)+\gamma+\frac{1}{2k}-\frac{1}{12k^2}+\frac{1}{120k^4}-\frac{1}{252k^6}+\frac{1}{240k^8}-\ldots)^4\]
显然我们先要解决下面和式的渐近估计
\[\sum_{k=1}^n \frac{\ln(k)^m}{k^p}\]
有谁能给出上面的和式渐近估计？ 以下是利用数学计算出来的结果。
当\(n=4\)时
\(\sum_{k=1}^n H_k^4\)
\[=(a_{00}+a_{01}\ln(n)+a_{02}\ln(n)^2+a_{03}\ln(n)^3+a_{04}\ln(n)^4)+(a_{10}+a_{11}\ln(n)+a_{12}\ln(n)^2+a_{13}\ln(n)^3+a_{14}\ln(n)^4)n+\frac{a_{20}+a_{21}\ln(n)+a_{22}\ln(n)^2+a_{23}\ln(n)^3}{n}+\frac{a_{30}+a_{31}\ln(n)+a_{32}\ln(n)^2+a_{33}\ln(n)^3}{n^2}+\]
\[\frac{a_{40}+a_{41}\ln(n)+a_{42}\ln(n)^2+a_{43}\ln(n)^3}{n^3}+\frac{a_{50}+a_{51}\ln(n)+a_{52}\ln(n)^2+a_{53}\ln(n)^3}{n^4}+\frac{a_{60}+a_{61}\ln(n)+a_{62}\ln(n)^2+a_{63}\ln(n)^3}{n^5}+\frac{a_{70}+a_{71}\ln(n)+a_{72}\ln(n)^2+a_{73}\ln(n)^3}{n^6}+\]
\[\frac{a_{80}+a_{81}\ln(n)+a_{82}\ln(n)^2+a_{83}\ln(n)^3}{n^7}+\frac{a_{90}+a_{91}\ln(n)+a_{92}\ln(n)^2+a_{93}\ln(n)^3}{n^8}+\ldots\]
\(a_{00}=?\)
\(a_{01}=0.769262067284738360\)
\(a_{02}=1.99906754284631205\)
\(a_{03}=2.30886265960613144\)
\(a_{04}=1\)
\(a_{10}=13.4867045896839190\)
\(a_{11}=-13.3756970607710970\)
\(a_{12}=7.07247956402791772\)
\(a_{13}=-1.69113734039386856\)
\(a_{14}=1\)
\(a_{20}=-1.42327876543675122\)
\(a_{21}=0.088673954137060519\)
\(a_{22}=2.3860783245076643\)
\(a_{23}=1.66666666666666667\)
\(a_{30}=0.435661713437849815\)
\(a_{31}=1.3984690708968799\)
\(a_{32}=0.922784335098467139\)
\(a_{33}=-0.333333333333333333\)
\(a_{40}=0.129623419937769641\)
\(a_{41}=-0.077906366759273163\)
\(a_{42}=-0.500611766843180047\)
\(a_{43}=0.0333333333333333333\)
\(a_{50}=-0.0615109855060221172\)
\(a_{51}=-0.168580902210767060\)
\(a_{52}=0.0993882331568199528\)
\(a_{53}=0.0333333333333333333\)
\(a_{60}=-0.0047558563025996941\)
\(a_{61}=0.0896557004029484970\)
\(a_{62}=0.0617992540523079590\)
\(a_{63}=-0.0158730158730158730\)
\(a_{70}=0.0176818009672201665\)
\(a_{71}=-0.028006913159222042\)
\(a_{72}=-0.0358197935667396600\)
\(a_{73}=-0.0158730158730158730\)
\(a_{80}=-0.00928934608136774666\)
\(a_{81}=-0.0156100566465372201\)
\(a_{82}=-0.0405638199295265316\)
\(a_{83}=0.016666666666666666\)
\(a_{90}=-0.00383966150441487355\)
\(a_{91}=0.0105106684897208053\)
\(a_{92}=0.0332457038799972779\)
\(a_{93}=0.0166666666666666667\)
当\(n=5\)时
\(\sum_{k=1}^n H_k^5\)
\[=(a_{00}+a_{01}\ln(n)+a_{02}\ln(n)^2+a_{03}\ln(n)^3+a_{04}\ln(n)^4+a_{05}\ln(n)^5)+(a_{10}+a_{11}\ln(n)+a_{12}\ln(n)^2+a_{13}\ln(n)^3+a_{14}\ln(n)^4+a_{15}\ln(n)^5)n+\frac{a_{20}+a_{21}\ln(n)+a_{22}\ln(n)^2+a_{23}\ln(n)^3+a_{24}\ln(n)^4 }{n}+\]
\[\frac{a_{30}+a_{31}\ln(n)+a_{32}\ln(n)^2+a_{33}\ln(n)^3)+a_{34}\ln(n)^4}{n^2}\]
\[\frac{a_{40}+a_{41}\ln(n)+a_{42}\ln(n)^2+a_{43}\ln(n)^3+a_{44}\ln(n)^4}{n^3}+\frac{a_{50}+a_{51}\ln(n)+a_{52}\ln(n)^2+a_{53}\ln(n)^3+a_{54}\ln(n)^4 }{n^4}+\frac{a_{60}+a_{61}\ln(n)+a_{62}\ln(n)^2+a_{63}\ln(n)^3+a_{64}\ln(n)^4}{n^5}+\frac{a_{70}+a_{71}\ln(n)+a_{72}\ln(n)^2+a_{73}\ln(n)^3)+a_{74}\ln(n)^4 }{n^6}\]
\[\frac{a_{80}+a_{81}\ln(n)+a_{82}\ln(n)^2+a_{83}\ln(n)^3+a_{84}\ln(n)^4 }{n^7}+\frac{ a_{90}+a_{91}\ln(n)+a_{92}\ln(n)^2+a_{93}\ln(n)^3+a_{94}\ln(n)^4}{n^8}+\ldots\]
\(a_{00}=?\)
\(a_{01}=0.555037644564109955\)
\(a_{02}=1.92315516821184590\)
\(a_{03}=3.33177923807718675\)
\(a_{04}=2.8860783245076643\)
\(a_{05}=1\)
\(a_{10}=-67.3694476638091040\)
\(a_{11}=67.4335229484195954\)
\(a_{12}=-33.4392426519277427\)
\(a_{13}=11.7874659400465296\)
\(a_{14}=-2.11392167549233570\)
\(a_{15}=1\)
\(a_{20}=-8.64802037947623567\)
\(a_{21}=-7.11639382718375607\)
\(a_{22}=0.22168488534265129\)
\(a_{23}=3.97679720751277384\)
\(a_{24}=2.08333333333333333\)
\(a_{30}=0.434535655005952312\)
\(a_{31}=2.17830856718924908\)
\(a_{32}=3.49617267724219977\)
\(a_{33}=1.53797389183077856\)
\(a_{34}=-0.416666666666666667\)
\(a_{40}=0.27622552978302117\)
\(a_{41}=0.648117099688848205\)
\(a_{42}=-0.194765916898182911\)
\(a_{43}=-0.834352944738633412\)
\(a_{44}=0.0416666666666666667\)
\(a_{50}=-0.009875749169700863\)
\(a_{51}=-0.307554927530110588\)
\(a_{52}=-0.421452255526917652\)
\(a_{53}=0.165647055261366588\)
\(a_{54}=0.0416666666666666667\)
\(a_{60}=-0.0390883404313376777\)
\(a_{61}=-0.023779281512998471\)
\(a_{62}=0.224139251007371243\)
\(a_{63}=0.102998756753846599\)
\(a_{64}=-0.0198412698412698413\)
\(a_{70}=0.0180435772862950085\)
\(a_{71}=0.088409004836100831\)
\(a_{72}=-0.0070017282898055105\)
\(a_{73}=-0.0596996559445661001\)
\(a_{74}=-0.0198412698412698413\)
\(a_{80}=0.0001363776836014963\)
\(a_{81}=-0.0464467304068387336\)
\(a_{82}=-0.0390251416163430506\)
\(a_{83}=-0.0676063665492108863\)
\(a_{84}=0.0208333333333333333\)
\(a_{90}=-0.0103355121764293795\)
\(a_{91}=-0.0191983075220743678\)
\(a_{92}=0.0262766712243020132\)
\(a_{93}=0.0554095064666621299\)
\(a_{94}=0.0208333333333333333\)
以上渐近表达式太复杂，不知有没有稍简洁的渐近公式？ 显然(H(n)-ln(n))^k更容易 mathe 发表于 2016-2-1 07:01
Sterling公式对数求导可以得出
Stirling展开式有
\(\ln\Gamma(x)=\frac{1}{2} \ln(2\pi)+(x-\frac{1}{2})\ln(x)+\sum_{n=1}^{\infty}\frac{B_{2n}}{2n(2n-1)x^{2n-1}}\)
两边求导有\(\frac{\Gamma'(x)}{\Gamma(x)}=\ln(x)+1-\frac{1}{2x}-\sum_{n=1}^{\infty}\frac{B_{2n}}{2nx^{2n}}\)
另外我们知道\(\ln(\Gamma(x+1))=\ln(x)+\ln(\Gamma(x))\)
求导得出\(\frac{\Gamma'(x+1)}{\Gamma(x+1)}=\frac{1}{x}+\frac{\Gamma'(x)}{\Gamma(x)}\)
可以得出\(\frac{\Gamma'(x)}{\Gamma(x)}=\frac{\Gamma'(x+n)}{\Gamma(x+n)}-\sum_{k=1}^n\frac{1}{x+k-1}\)
$x=1$代入可以得出\(H_n=\frac{\Gamma'(1+n)}{\Gamma(1+n)}+\gamma\)
然后将公式\(\frac{\Gamma'(x)}{\Gamma(x)}=\ln(x)+1-\frac{1}{2x}-\sum_{n=1}^{\infty}\frac{B_{2n}}{2nx^{2n}}\)中用$x=n+1$代入即可得到$H_n$的渐进式 :*-^，漏掉了这么精彩的帖子呀，顶起来！ wayne 发表于 2018-1-27 13:01
，漏掉了这么精彩的帖子呀，顶起来！
漏掉了这么精彩的帖子呀，顶起来！
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