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标题: ln1+2ln2+3ln3+...+nln(n) 有限项的和 [打印本页]

作者: 王守恩 时间: 2018-6-9 08:58
标题: ln1+2ln2+3ln3+...+nln(n) 有限项的和
求助：ln1+2ln2+3ln3+4ln4+...+nln(n) 有限项的和

作者: qingjiao 时间: 2018-6-9 17:22

=n∑ln(n)-∑ln1-∑ln2-∑ln3-...-∑ln(n-2)-∑ln(n-1)

然后楼主逐个应用：∑ln(m)≈mlnm-m+(lnm)/2+(ln2π)/2+1/12m 的公式慢慢化简吧

作者: qingjiao 时间: 2018-6-9 20:11
不好意思，楼上的算法可能只有一个循环的结果，建议通过ln(1+x)的展开式求和。

ln(n-r)=ln[(n-r)/n]+ln(n)=ln(1-r/n)+ln(n)=-r/n-(1/2)(r/n)^2-(1/3)(r/n)^3-...+ln(n)，r=1,2,3,..., n-1

这样可以将每个ln(m)=ln(n-r)项展开，楼主再慢求和简化吧。

作者: 282842712474 时间: 2018-6-11 19:35
https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula

$$\sum _{i=m}^{n}f(i)-\int _{m}^{n}f(x)~{\rm {d}}x={\frac {f(m)+f(n)}{2}}+{\frac {1}{6}}{\frac {f'(n)-f'(m)}{2!}}-{\frac {1}{30}}{\frac {f'''(n)-f'''(m)}{4!}}+\cdots +B_{2k}{\frac {f^{(2k-1)}(n)-f^{(2k-1)}(m)}{(2k)!}}+R_{2k}.$$

作者: 王守恩 时间: 2018-6-12 14:51

282842712474 发表于 2018-6-11 19:35
https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula

$$\sum _{i=m}^{n}f(i)-\int _{m}^{n} ...

   求助。我们可以这样说吗？

   \(\D\lim_{\theta\to\infty}\frac{n\ln(1)+n\ln(2)+n\ln(3)+...+n\ln(n)}{1\ln(1)+2\ln(2)+3\ln(3)+...+n\ln(n)}=2\)

作者: mathe 时间: 2018-6-12 17:03
https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem
设\((a_n)_{n\ge 1}\)和\((b_n)_{n\ge 1}\) 是两个实数序列。如果\((b_n)_{n\ge 1}\) 严格单调增而且发散，而且极限\(\lim_{n->\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=l\)
那么必然有\(\lim_{n->\infty}\frac{a_n}{b_n}=l\)
所以取\(a_n=n\ln(1)+n\ln(2)+...+n\ln(n), b_n=1\ln(1)+2\ln(2)+...+n\ln(n)\)
于是\((b_n)_{n\ge 1}\)严格单调增，
而\(\lim_{n->\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n->\infty}\frac{\ln(1)+\ln(2)+...+\ln(n)+(n+1)\ln(n+1)}{(n+1)\ln(n+1)}=1+\lim_{n->\infty}\frac{\ln(1)+\ln(2)+...+\ln(n)}{(n+1)\ln(n+1)}=1+\lim_{n->\infty}\frac{\ln(n!)}{(n+1)\ln(n+1)}\)
根据Stirling公式$\ln(\sqrt(2pi n))+n\ln(n)-n\lt \ln(n!) \lt\ln(\sqrt(2pi n))+n\ln(n)-n+1/{12n}$
所以\(\lim_{n->\infty}\frac{\ln(n!)}{(n+1)\ln(n+1)}=1\)
也就是\(\lim_{n->\infty}\frac{a_n}{b_n}=\lim_{n->\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=2\)

作者: 王守恩 时间: 2018-6-16 10:17

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 ...

谢谢mathe！照葫芦画瓢，我还是画不好。下面的算式能成立吗？

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

作者: 王守恩 时间: 2018-6-19 19:57

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 ...

7楼的算式有问题，我来改一下！

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

作者: 王守恩 时间: 2018-6-22 06:08
本帖最后由王守恩于 2018-6-22 06:46 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 ...

谢谢mathe！求助！我是算不好的！

\(\ \ \ \D 若\lim_{n\to\infty}a_{n}=a，a_{n}\gt 0 (n=1,2,3,\cdots)，计算极限：\)

\(\ \ \ \ \ \ \D \lim_{n\to\infty}\lim_{x\to0}\left(\frac{a_{1}^x+a_{2}^x+a_{3}^x+\cdots+a_{n}^x}{n}\right)^{\frac{1}{x}}\)

作者: mathe 时间: 2018-6-22 13:19
\(\lim_{x\to0}\left(\frac{a_{1}^x+a_{2}^x+a_{3}^x+\cdots+a_{n}^x}{n}\right)^{\frac{1}{x}}=\sqrt[n]{a_1a_2a_3...a_n}\)
所以9#的答案很显然。8#的也没有问题，其实只要分子分母都和$\ln^{n+1}(n)$比较即可

作者: 王守恩 时间: 2018-6-26 17:20

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” 。

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

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

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

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

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

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

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

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

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

   \(......\)

作者: 王守恩 时间: 2018-6-27 12:28

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” 。

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

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

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

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

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

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

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

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

作者: 王守恩 时间: 2018-7-2 13:25
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 ...

   \(我们有：\)

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

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

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

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

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

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

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

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

作者: wayne 时间: 2018-7-2 15:42

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$x=-0.29590500557521395564723783108304803394867416605194782899479943464744358207245187792168714360217155879828829399693624245615675374124923920698381661562339315906004649448542384429238724232806000084284327$

作者: 王守恩 时间: 2018-7-3 19:53

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？

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

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

作者: wayne 时间: 2018-7-3 20:39

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）

作者: 王守恩 时间: 2019-2-12 08:56
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 ...

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

作者: wayne 时间: 2019-2-13 11:03

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

作者: 王守恩 时间: 2019-2-15 16:58
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)

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

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

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

作者: 王守恩 时间: 2019-3-3 16:11
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)

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

\(\D\cdots\cdots\)

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

\(\D\cdots\cdots\)

作者: 王守恩 时间: 2019-4-5 09:06
本帖最后由王守恩于 2019-4-5 09:10 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 ...

谢谢wayne！左边不好算，右边好算。

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

            \(\D\cdots\cdots\)

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

            \(\D\cdots\cdots\)

作者: 王守恩 时间: 2019-4-8 06:30

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？

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

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