ln1+2ln2+3ln3+...+nln(n) 有限项的和

王守恩

求助：ln1+2ln2+3ln3+4ln4+...+nln(n)  有限项的和

王守恩

王守恩 发表于 2019-4-5 09:06
谢谢wayne！ 左边不好算，右边好算。

我们可以这样说吗？

\(\D\sum_{k=2}^{n}\ \zeta k\ =n\)

王守恩

wayne 发表于 2018-7-3 20:39
指数为负的时候，级数是发散的。通过解析延拓可扩大定义范围（https://en.wikipedia.org/wiki/Analytic ...

谢谢wayne！ 左边不好算，右边好算。


\(公式(1)：\D\sum_{k=1}^{n}\frac{1}{\sqrt[A]{k}}=\frac{A}{A-1}\cdot  n^{\frac{A-1}{A}}+\zeta\bigg(\frac{1}{A}\bigg)\ 譬如：\)
              \(\D\sum_{k=1}^{n}\frac{1}{\sqrt[2]{k}}=\frac{2}{1}\cdot  n^{\frac{1}{2}}+\zeta\bigg(\frac{1}{2}\bigg)\)
              \(\D\sum_{k=1}^{n}\frac{1}{\sqrt[3]{k}}=\frac{3}{2}\cdot  n^{\frac{2}{3}}+\zeta\bigg(\frac{1}{3}\bigg)\)
              \(\D\sum_{k=1}^{n}\frac{1}{\sqrt[4]{k}}=\frac{4}{3}\cdot  n^{\frac{3}{4}}+\zeta\bigg(\frac{1}{4}\bigg)\)
              \(\D\sum_{k=1}^{n}\frac{1}{\sqrt[5]{k}}=\frac{5}{4}\cdot  n^{\frac{4}{5}}+\zeta\bigg(\frac{1}{5}\bigg)\)
              \(\D\sum_{k=1}^{n}\frac{1}{\sqrt[6]{k}}=\frac{6}{5}\cdot  n^{\frac{5}{6}}+\zeta\bigg(\frac{1}{6}\bigg)\)
              \(\D\sum_{k=1}^{n}\frac{1}{\sqrt[7]{k}}=\frac{7}{6}\cdot  n^{\frac{6}{7}}+\zeta\bigg(\frac{1}{7}\bigg)\)
            
              \(\D\cdots\cdots\)

\(公式(2)：\D\sum_{k=1}^{n}\frac{1}{k^{A}}=\frac{A^{-1}}{A^{-1}-1}\cdot  n^{\frac{A^{-1}-1}{A^{-1}}}+\zeta\bigg(A\bigg)\ 譬如：\)
               \(\D\sum_{k=1}^{n}\frac{1}{k^{2}}=\frac{2^{-1}}{2^{-1}-1}\cdot  n^{\frac{2^{-1}-1}{2^{-1}}}+\zeta\bigg(2\bigg)\)
               \(\D\sum_{k=1}^{n}\frac{1}{k^{3}}=\frac{3^{-1}}{3^{-1}-1}\cdot  n^{\frac{3^{-1}-1}{3^{-1}}}+\zeta\bigg(3\bigg)\)
               \(\D\sum_{k=1}^{n}\frac{1}{k^{4}}=\frac{4^{-1}}{4^{-1}-1}\cdot  n^{\frac{4^{-1}-1}{4^{-1}}}+\zeta\bigg(4\bigg)\)
               \(\D\sum_{k=1}^{n}\frac{1}{k^{5}}=\frac{5^{-1}}{5^{-1}-1}\cdot  n^{\frac{5^{-1}-1}{5^{-1}}}+\zeta\bigg(5\bigg)\)
               \(\D\sum_{k=1}^{n}\frac{1}{k^{6}}=\frac{6^{-1}}{6^{-1}-1}\cdot  n^{\frac{6^{-1}-1}{6^{-1}}}+\zeta\bigg(6\bigg)\)
               \(\D\sum_{k=1}^{n}\frac{1}{k^{7}}=\frac{7^{-1}}{7^{-1}-1}\cdot  n^{\frac{7^{-1}-1}{7^{-1}}}+\zeta\bigg(7\bigg)\)
           
               \(\D\cdots\cdots\)

王守恩

烦请会电脑的网友用数据验算一下(不用证明)：
下面的不等号是不是成立，或者能举个反例更好，不胜感谢！
   
   A(0) < T(n) < B(0)
   A(1) < T(n) < B(1)
   A(2) < T(n) < B(2)
   A(3) < T(n) < B(3)
   A(4) < T(n) < B(4)
   A(5) < T(n) < B(5)
   A(6) < T(n) < B(6)
   A(7) < T(n) < B(7)
   A(8) < T(n) < B(8)
   A(9) < T(n) < B(9)


\(\D T(n)=\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{4}+\cdots\sqrt{\frac{1}{n}}}}}}\)
\(\D A(0)=\sqrt{\frac{1+1}{1}}\)
\(\D A(1)=\sqrt{1+\sqrt{\frac{2+1}{2}}}\)
\(\D A(2)=\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{3+1}{3}}}}\)
\(\D A(3)=\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{4+1}{4}}}}}\)
\(\D A(4)=\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{4}+\sqrt{\frac{5+1}{5}}}}}}\)

\(\D\cdots\cdots\)

\(\D B(0)=\sqrt{\frac{1+1}{1}}\)
\(\D B(1)=\sqrt{1+\sqrt{\frac{2+2}{2}}}\)
\(\D B(2)=\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{3+2}{3}}}}\)
\(\D B(3)=\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{4+2}{4}}}}}\)
\(\D B(4)=\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{4}+\sqrt{\frac{5+2}{5}}}}}}\)

\(\D\cdots\cdots\)

王守恩

wayne 发表于 2019-2-13 11:03

wayne ！我们可以这样说吗？

   A(0) < T(n) < B(0)
   A(1) < T(n) < B(1)
   A(2) < T(n) < B(2)
   A(3) < T(n) < B(3)
   A(4) < T(n) < B(4)
   A(5) < T(n) < B(5)
   A(6) < T(n) < B(6)
   A(7) < T(n) < B(7)
   A(8) < T(n) < B(8)
   A(9) < T(n) < B(9)

\(\D T(n)=\lim_{n\to\infty}\sqrt{1+\sqrt{\frac{1}{2^{2}}+\sqrt{\frac{1}{3^{2}}+\sqrt{\frac{1}{4^{2}}+\sqrt{\frac{1}{5^{2}}+\cdots\sqrt{\frac{1}{n^{2}}}}}}}}\)

\(A(0)=\sqrt{\frac{1^{2}+1}{1^{2}}}\)
\(A(1)=\sqrt{1+\sqrt{\frac{2^{2}+1}{2^{2}}}}\)
\(A(2)=\sqrt{1+\sqrt{\frac{1}{2^{2}}+\sqrt{\frac{3^{2}+1}{3^{2}}}}}\)
\(A(3)=\sqrt{1+\sqrt{\frac{1}{2^{2}}+\sqrt{\frac{1}{3^{2}}+\sqrt{\frac{4^{2}+1}{4^{2}}}}}}\)
\(A(4)=\sqrt{1+\sqrt{\frac{1}{2^{2}}+\sqrt{\frac{1}{3^{2}}+\sqrt{\frac{1}{4^{2}}+\sqrt{\frac{5^{2}+1}{5^{2}}}}}}}\)
\(A(5)=\sqrt{1+\sqrt{\frac{1}{2^{2}}+\sqrt{\frac{1}{3^{2}}+\sqrt{\frac{1}{4^{2}}+\sqrt{\frac{1}{5^{2}}+\sqrt{\frac{6^{2}+1}{6^{2}}}}}}}}\)
\(\cdots\cdots\cdots\)

\(B(0)=\sqrt{\frac{1^{2}+2}{1^{2}}}\)
\(B(1)=\sqrt{1+\sqrt{\frac{2^{2}+2}{2^{2}}}}\)
\(B(2)=\sqrt{1+\sqrt{\frac{1}{2^{2}}+\sqrt{\frac{3^{2}+2}{3^{2}}}}}\)
\(B(3)=\sqrt{1+\sqrt{\frac{1}{2^{2}}+\sqrt{\frac{1}{3^{2}}+\sqrt{\frac{4^{2}+2}{4^{2}}}}}}\)
\(B(4)=\sqrt{1+\sqrt{\frac{1}{2^{2}}+\sqrt{\frac{1}{3^{2}}+\sqrt{\frac{1}{4^{2}}+\sqrt{\frac{5^{2}+2}{5^{2}}}}}}}\)
\(B(5)=\sqrt{1+\sqrt{\frac{1}{2^{2}}+\sqrt{\frac{1}{3^{2}}+\sqrt{\frac{1}{4^{2}}+\sqrt{\frac{1}{5^{2}}+\sqrt{\frac{6^{2}+2}{6^{2}}}}}}}}\)
\(\cdots\cdots\cdots\)

wayne

王守恩 发表于 2019-2-12 08:56
\[\D\lim_{n\to\infty}\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\sqrt{\frac{1}{4^2}+\sqrt{\ ...

1. N[Fold[Sqrt[1/#2^2 + #1] &, 0, Range[1000, 1, -1]], 1000]

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1. 1.466723564171902713805754180665991105424967782836996958138363948238662157825638193253678988440041823285098811661578907809636230110019362946839523952280872210431563731746173882668211511659893991978352969167244687910659590285286164978285060798241893293624489066655657914199142453463449288940109105689926167476215037719621877523422458049454780953615199985031704265238811016113804745912543170048593570180902580002328537189659270070821725124775539810161611356761104725709790478713151827341488444998874794873133964421078833417402680718840428684827921012169685429302057893375700135850055995386693494947332290995845331208516829943422346499699162074433432270484798996200448812335376014675750238773986295930225524101553606958142009902158226776087686116402852803190621766721107205487371396506541784100719696359463496210928872450126856935529303536091448256591409003409636656069181972070979789069101155830075644014060440874082817013633422172154218276819156811152526266423766040130296104650654487568789649478041162...

复制代码

王守恩

wayne 发表于 2018-7-3 20:39
指数为负的时候，级数是发散的。通过解析延拓可扩大定义范围（https://en.wikipedia.org/wiki/Analytic ...

\[\D\lim_{n\to\infty}\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\sqrt{\frac{1}{4^2}+\sqrt{\frac{1}{5^2}+\cdots\cdots\sqrt{\frac{1}{n^2}}}}}}}=?\]

wayne

王守恩 发表于 2018-7-3 19:53
谢谢wayne！还是请教！第1个算式是对的！第2个算式左右不相等？

\(\D\lim_{n\to\infty}\frac{1}{1^{1. ...

指数为负的时候，级数是发散的。通过解析延拓可扩大定义范围（https://en.wikipedia.org/wiki/Analytic_continuation）

王守恩

wayne 发表于 2018-7-2 15:42
就是计算黎曼zeta函数的不动点．
一个是$x = 1.833772651680271396245648589441523592180978518800993 ...

谢谢wayne！还是请教！第1个算式是对的！第2个算式左右不相等？

\(\D\lim_{n\to\infty}\frac{1}{1^{1.83377}}+\frac{1}{2^{1.83377}}+\frac{1}{3^{1.83377}}+\frac{1}{4^{1.83377}}+\cdots+\frac{1}{n^{1.83377}}=1.83377\)

\(\D\lim_{n\to\infty}\frac{1}{1^{-0.2959}}+\frac{1}{2^{-0.2959}}+\frac{1}{3^{-0.2959}}+\frac{1}{4^{-0.2959}}+\cdots+\frac{1}{n^{-0.2959}}=-0.2959\)

wayne

王守恩 发表于 2018-7-2 13:25
\(我们有：\)

    \(\D\lim_{n\to\infty}\frac{1}{1^{1.00000}}+\frac{1}{2^{1.00000}}+\frac ...

就是计算黎曼zeta函数的不动点．
一个是$x = 1.8337726516802713962456485894415235921809785188009933371940375600980726720056881390347430959755443918066045153558760309025588985749154100738953098696620795133064336889892730397119789328429160367753883$
一个是$x=-0.29590500557521395564723783108304803394867416605194782899479943464744358207245187792168714360217155879828829399693624245615675374124923920698381661562339315906004649448542384429238724232806000084284327$

王守恩

mathe 发表于 2018-6-12 17:03
https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem
设\((a_n)_{n\ge 1}\)和\((b_n)_{n\ge ...

      \(我们有：\)

    \(\D\lim_{n\to\infty}\frac{1}{1^{1.00000}}+\frac{1}{2^{1.00000}}+\frac{1}{3^{1.00000}}+\frac{1}{4^{1.00000}}+\cdots+\frac{1}{n^{1.00000}}=\infty\)

    \(\D\lim_{n\to\infty}\frac{1}{1^{1.00001}}+\frac{1}{2^{1.00001}}+\frac{1}{3^{1.00001}}+\frac{1}{4^{1.00001}}+\cdots+\frac{1}{n^{1.00001}}=100001\)

    \(\D\lim_{n\to\infty}\frac{1}{1^{1.00010}}+\frac{1}{2^{1.00010}}+\frac{1}{3^{1.00010}}+\frac{1}{4^{1.00010}}+\cdots+\frac{1}{n^{1.00010}}=10000.6\)

    \(\D\lim_{n\to\infty}\frac{1}{1^{1.00100}}+\frac{1}{2^{1.00100}}+\frac{1}{3^{1.00100}}+\frac{1}{4^{1.00100}}+\cdots+\frac{1}{n^{1.00100}}=1000.58\)

    \(\D\lim_{n\to\infty}\frac{1}{1^{1.01000}}+\frac{1}{2^{1.01000}}+\frac{1}{3^{1.01000}}+\frac{1}{4^{1.01000}}+\cdots+\frac{1}{n^{1.01000}}=100.578\)

    \(\D\lim_{n\to\infty}\frac{1}{1^{1.10000}}+\frac{1}{2^{1.10000}}+\frac{1}{3^{1.10000}}+\frac{1}{4^{1.10000}}+\cdots+\frac{1}{n^{1.10000}}=10.5844\)

       \(\cdots\cdots\cdots\cdots\)

      \(求助：当\D\lim_{n\to\infty}\frac{1}{1^{x}}+\frac{1}{2^{x}}+\frac{1}{3^{x}}+\frac{1}{4^{x}}+\cdots+\frac{1}{n^{x}}=\D x\ 时，\D x=\ ?\)