ln1+2ln2+3ln3+...+nln(n) 有限项的和

王守恩 · 发表于 2018-6-26 17:20:48

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” 。

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

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

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

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

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

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

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

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

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

   $......$

王守恩 · 发表于 2018-6-27 12:28:53

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” 。

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

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

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

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

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

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

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

   $\cdots\cdots\cdots\cdots\cdots$

王守恩 · 发表于 2018-7-2 13:25:28

本帖最后由王守恩于 2018-7-2 13:31 编辑

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 ...

   $我们有：$

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

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

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

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

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

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

   $\cdots\cdots\cdots\cdots$

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

wayne · 发表于 2018-7-2 15:42:41

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 · 发表于 2018-7-3 19:53:58

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 · 发表于 2018-7-3 20:39:32

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 · 发表于 2019-2-12 08:56:16

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 ...

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

wayne · 发表于 2019-2-13 11:03:44

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{\ ...

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

复制代码

1.466723564171902713805754180665991105424967782836996958138363948238662157825638193253678988440041823285098811661578907809636230110019362946839523952280872210431563731746173882668211511659893991978352969167244687910659590285286164978285060798241893293624489066655657914199142453463449288940109105689926167476215037719621877523422458049454780953615199985031704265238811016113804745912543170048593570180902580002328537189659270070821725124775539810161611356761104725709790478713151827341488444998874794873133964421078833417402680718840428684827921012169685429302057893375700135850055995386693494947332290995845331208516829943422346499699162074433432270484798996200448812335376014675750238773986295930225524101553606958142009902158226776087686116402852803190621766721107205487371396506541784100719696359463496210928872450126856935529303536091448256591409003409636656069181972070979789069101155830075644014060440874082817013633422172154218276819156811152526266423766040130296104650654487568789649478041162...

复制代码

王守恩 · 发表于 2019-2-15 16:58:12

本帖最后由王守恩于 2019-2-15 17:44 编辑

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)

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

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

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

王守恩 · 发表于 2019-3-3 16:11:44

本帖最后由王守恩于 2019-3-3 16:14 编辑

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)

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

$\D\cdots\cdots$

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

$\D\cdots\cdots$

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[求助] ln1+2ln2+3ln3+...+nln(n) 有限项的和

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