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

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？

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

         

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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？


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

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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！


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

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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！

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

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

王守恩

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

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

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 ...

      \(我们有：\)

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

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

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

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

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

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

       \(\cdots\cdots\cdots\cdots\)

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

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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？

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

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

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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 ...

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