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楼主: 王守恩

[求助] ln1+2ln2+3ln3+...+nln(n) 有限项的和

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 楼主| 发表于 2018-6-26 17:20:48 | 显示全部楼层
mathe 发表于 2018-6-22 13:19
\(\lim_{x\to0}\left(\frac{a_{1}^x+a_{2}^x+a_{3}^x+\cdots+a_{n}^x}{n}\right)^{\frac{1}{x}}=\sqrt[n]{a ...

所有数相加的和是  “1” 。


     \(\D\frac{1}{02!}+\frac{1}{03!}+\frac{1}{04!}+\frac{1}{05!}+\frac{1}{06!}+\frac{1}{07!}+\frac{1}{08!}+\frac{1}{09!}+\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\cdots\)

     \(\D\frac{1}{03!}+\frac{1}{04!}+\frac{1}{05!}+\frac{1}{06!}+\frac{1}{07!}+\frac{1}{08!}+\frac{1}{09!}+\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\frac{1}{13!}+\cdots\)

     \(\D\frac{1}{04!}+\frac{1}{05!}+\frac{1}{06!}+\frac{1}{07!}+\frac{1}{08!}+\frac{1}{09!}+\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\frac{1}{13!}+\frac{1}{14!}+\cdots\)

     \(\D\frac{1}{05!}+\frac{1}{06!}+\frac{1}{07!}+\frac{1}{08!}+\frac{1}{09!}+\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\frac{1}{13!}+\frac{1}{14!}+\frac{1}{15!}+\cdots\)

     \(\D\frac{1}{06!}+\frac{1}{07!}+\frac{1}{08!}+\frac{1}{09!}+\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\frac{1}{13!}+\frac{1}{14!}+\frac{1}{15!}+\frac{1}{16!}+\cdots\)

     \(\D\frac{1}{07!}+\frac{1}{08!}+\frac{1}{09!}+\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\frac{1}{13!}+\frac{1}{14!}+\frac{1}{15!}+\frac{1}{16!}+\frac{1}{17!}+\cdots\)

     \(\D\frac{1}{08!}+\frac{1}{09!}+\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\frac{1}{13!}+\frac{1}{14!}+\frac{1}{15!}+\frac{1}{16!}+\frac{1}{17!}+\frac{1}{18!}+\cdots\)

     \(\D\frac{1}{09!}+\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\frac{1}{13!}+\frac{1}{14!}+\frac{1}{15!}+\frac{1}{16!}+\frac{1}{17!}+\frac{1}{18!}+\frac{1}{19!}+\cdots\)

     \(\D\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\frac{1}{13!}+\frac{1}{14!}+\frac{1}{15!}+\frac{1}{16!}+\frac{1}{17!}+\frac{1}{18!}+\frac{1}{19!}+\frac{1}{20!}+\cdots\)

     \(......\)




毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2018-6-27 12:28:53 | 显示全部楼层
王守恩 发表于 2018-6-26 17:20
所有数相加的和是  “1” 。


所有数相加的和是  “1” 。




      \(\D\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9^2}+\frac{1}{10^2}+\frac{1}{11^2}+\frac{1}{12^2}+\cdots\)

      \(\D\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+\frac{1}{8^3}+\frac{1}{9^3}+\frac{1}{10^3}+\frac{1}{11^3}+\frac{1}{12^3}+\cdots\)

      \(\D\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}+\frac{1}{8^4}+\frac{1}{9^4}+\frac{1}{10^4}+\frac{1}{11^4}+\frac{1}{12^4}+\cdots\)

      \(\D\frac{1}{2^5}+\frac{1}{3^5}+\frac{1}{4^5}+\frac{1}{5^5}+\frac{1}{6^5}+\frac{1}{7^5}+\frac{1}{8^5}+\frac{1}{9^5}+\frac{1}{10^5}+\frac{1}{11^5}+\frac{1}{12^5}+\cdots\)

      \(\D\frac{1}{2^6}+\frac{1}{3^6}+\frac{1}{4^6}+\frac{1}{5^6}+\frac{1}{6^6}+\frac{1}{7^6}+\frac{1}{8^6}+\frac{1}{9^6}+\frac{1}{10^6}+\frac{1}{11^6}+\frac{1}{12^6}+\cdots\)

      \(\D\frac{1}{2^7}+\frac{1}{3^7}+\frac{1}{4^7}+\frac{1}{5^7}+\frac{1}{6^7}+\frac{1}{7^7}+\frac{1}{8^7}+\frac{1}{9^7}+\frac{1}{10^7}+\frac{1}{11^7}+\frac{1}{12^7}+\cdots\)

      \(\D\frac{1}{2^8}+\frac{1}{3^8}+\frac{1}{4^8}+\frac{1}{5^8}+\frac{1}{6^8}+\frac{1}{7^8}+\frac{1}{8^8}+\frac{1}{9^8}+\frac{1}{10^8}+\frac{1}{11^8}+\frac{1}{12^8}+\cdots\)

       \(\cdots\cdots\cdots\cdots\cdots\)




毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2018-7-2 13:25:28 | 显示全部楼层
本帖最后由 王守恩 于 2018-7-2 13:31 编辑


      \(我们有:\)

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

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

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

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

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

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

       \(\cdots\cdots\cdots\cdots\)

      \(求助:当\D\lim_{n\to\infty}\frac{1}{1^{x}}+\frac{1}{2^{x}}+\frac{1}{3^{x}}+\frac{1}{4^{x}}+\cdots+\frac{1}{n^{x}}=\D x\ 时,\D x=\ ?\)  
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2018-7-2 15:42:41 | 显示全部楼层
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$x=-0.29590500557521395564723783108304803394867416605194782899479943464744358207245187792168714360217155879828829399693624245615675374124923920698381661562339315906004649448542384429238724232806000084284327$
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2018-7-3 19:53:58 | 显示全部楼层
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?

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

\(\D\lim_{n\to\infty}\frac{1}{1^{-0.2959}}+\frac{1}{2^{-0.2959}}+\frac{1}{3^{-0.2959}}+\frac{1}{4^{-0.2959}}+\cdots+\frac{1}{n^{-0.2959}}=-0.2959\)
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2018-7-3 20:39:32 | 显示全部楼层
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

点评

谢谢wayne!  发表于 2018-7-3 20:57
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 5 天前 | 显示全部楼层
本帖最后由 王守恩 于 2019-2-12 09:03 编辑
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 ...


\[\D\lim_{n\to\infty}\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\sqrt{\frac{1}{4^2}+\sqrt{\frac{1}{5^2}+\cdots\cdots\sqrt{\frac{1}{n^2}}}}}}}=?\]
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 4 天前 | 显示全部楼层
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...
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点评

谢谢wayne!  发表于 3 天前
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 前天 16:58 | 显示全部楼层
本帖最后由 王守恩 于 2019-2-15 17:44 编辑


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)

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

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

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



毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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