伟大的 "e"
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n...+n^n+(n+1)^n}{1+2^n+3^n...+(n-1)^n+n^n}=\D e\)\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n+0)^n+(n+1)^n}{1+2^n+3^n+...+(n−1)^n+n^n}=e^1\)
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n+1)^n+(n+2)^n}{1+2^n+3^n+...+(n−1)^n+n^n}=e^2\)
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n+2)^n+(n+3)^n}{1+2^n+3^n+...+(n−1)^n+n^n}=e^3\)
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n+3)^n+(n+4)^n}{1+2^n+3^n+...+(n−1)^n+n^n}=e^4\)
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n+4)^n+(n+5)^n}{1+2^n+3^n+...+(n−1)^n+n^n}=e^5\) 王守恩 发表于 2018-5-25 11:34
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n+0)^n+(n+1)^n}{1+2^n+3^n+...+(n−1)^n+n^n}=e^1\)
...
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n-1)^n+n^n}{1+2^n+3^n+...+(n+0)^n+(n+1)^n}=\frac{1}{e^1}\)
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n-1)^n+n^n}{1+2^n+3^n+...+(n+1)^n+(n+2)^n}=\frac{1}{e^2}\)
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n-1)^n+n^n}{1+2^n+3^n+...+(n+2)^n+(n+3)^n}=\frac{1}{e^3}\)
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n-1)^n+n^n}{1+2^n+3^n+...+(n+3)^n+(n+4)^n}=\frac{1}{e^4}\)
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n-1)^n+n^n}{1+2^n+3^n+...+(n+4)^n+(n+5)^n}=\frac{1}{e^5}\)
王守恩 发表于 2018-5-25 11:34
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n+0)^n+(n+1)^n}{1+2^n+3^n+...+(n−1)^n+n^n}=e^1\)
...
如何证明？ 本帖最后由 王守恩 于 2018-5-26 16:07 编辑
王守恩 发表于 2018-5-25 18:05
\(\D\lim_{n\to\infty}\frac{1+2^n+3^n+...+(n-1)^n+n^n}{1+2^n+3^n+...+(n+0)^n+(n+1)^n} ...
我们有
\(\D\lim_{n\to\infty}\frac{1^1+2^1+3^1+4^1+...+n^1}{n^2}=\frac{1}{2}\)
\(\D\lim_{n\to\infty}\frac{1^2+2^2+3^2+4^2+...+n^2}{n^3}=\frac{1}{3}\)
\(\D\lim_{n\to\infty}\frac{1^3+2^3+3^3+4^3+...+n^3}{n^4}=\frac{1}{4}\)
\(\D\lim_{n\to\infty}\frac{1^4+2^4+3^4+4^4+...+n^4}{n^5}=\frac{1}{5}\)
\(.........\)
同理可得
\(\D\lim_{n\to\infty}\frac{1^{e-2}+2^{e-2}+3^{e-2}+...+n^{e-2}}{n^{e-1}}=\frac{1}{e-1}\)
\(\D\lim_{n\to\infty}\frac{1^{e-1}+2^{e-1}+3^{e-1}+...+n^{e-1}}{n^{e-0}}=\frac{1}{e-0}\)
\(\D\lim_{n\to\infty}\frac{1^{e-0}+2^{e-0}+3^{e-0}+...+n^{e-0}}{n^{e+1}}=\frac{1}{e+1}\)
\(\D\lim_{n\to\infty}\frac{1^{e+1}+2^{e+1}+3^{e+1}+...+n^{e+1}}{n^{e+2}}=\frac{1}{e+2}\)
\(.........\)
王守恩 发表于 2018-5-26 16:00
我们有
\(\D\lim_{n\to\infty}\frac{1^1+2^1+3^1+4^1+...+n^1}{n^2}=\frac{1}{2} ...
\(\D\lim_{n\to\infty}\frac{1+2^{e+3}+3^{e+3}+...+n^{e+3}}{(1+2^{e-3}+3^{e-3}+...+n^{e-3})×n^6}=\frac{e -2}{e+4}\)
\(\D\lim_{n\to\infty}\frac{1+2^{e+1}+3^{e+1}+...+n^{e+1}}{(1+2^{e-3}+3^{e-3}+...+n^{e-3})×n^4}=\frac{e -2}{e+2}\)
\(\D\lim_{n\to\infty}\frac{1+2^{e}+3^{e}+...+n^{e}}{(1+2^{e-2}+3^{e-2}+...+n^{e-2})×n^2}=\frac{e -1}{e+1}\)
\(\D\lim_{n\to\infty}\frac{(1+2^{e-1}+3^{e-1}+...+n^{e-1})×n}{1+2^{e}+3^{e}+...+n^{e}}=\frac{e+1}{e}\)
\(\D\lim_{n\to\infty}\frac{(1+2^{e-2}+3^{e-2}+...+n^{e-2})×n^3}{1+2^{e+1}+3^{e+1}+...+n^{e+1}}=\frac{e+2}{e -1}\)
\(\D\lim_{n\to\infty}\frac{(1+2^{e-2}+3^{e-2}+...+n^{e-2})×n^5}{1+2^{e+3}+3^{e+3}+...+n^{e+3}}=\frac{e+4}{e -1}\)
\(\D\lim_{n\to\infty}\frac{(1+2^{e-3}+3^{e-3}+...+n^{e-3})×n^7}{1+2^{e+4}+3^{e+4}+...+n^{e+4}}=\frac{e+5}{e -2}\) 王守恩 发表于 2018-5-28 09:34
\(\D\lim_{n\to\infty}\frac{1+2^{e+3}+3^{e+3}+...+n^{e+3}}{(1+2^{e-3}+3^{e-3}+...+n^{e-3})×n^6}=\f ...
王守恩 发表于 2018-5-26 16:00
我们有
\(\D\lim_{n\to\infty}\frac{1^1+2^1+3^1+4^1+...+n^1}{n^2}=\frac{1}{2} ...
\(\D\lim_{n\to\infty}\frac{1+2+3+4+...+n}{(1+1+1+1+...+1)×n}=\frac{1}{2}\)
\(\D\lim_{n\to\infty}\frac{1+2^2+3^2+4^2+...+n^2}{(1+2+3+4+...+n)×n}=\frac{2}{3}\)
\(\D\lim_{n\to\infty}\frac{1+2^3+3^3+4^3+...+n^3}{(1+2^2+3^2+4^2+...+n^2)×n}=\frac{3}{4}\)
\(\D\lim_{n\to\infty}\frac{1+2^4+3^4+4^4+...+n^4}{(1+2^3+3^3+4^3+...+n^3)×n}=\frac{4}{5}\)
\(\D\lim_{n\to\infty}\frac{1+2^5+3^5+4^5+...+n^5}{(1+2^4+3^4+4^4+...+n^4)×n}=\frac{5}{6}\)
\(\D\lim_{n\to\infty}\frac{1+2^6+3^6+4^6+...+n^6}{(1+2^5+3^5+4^5+...+n^5)×n}=\frac{6}{7}\)
\(\D\lim_{n\to\infty}\frac{1+2^7+3^7+4^7+...+n^7}{(1+2^6+3^6+4^6+...+n^6)×n}=\frac{7}{8}\)
王守恩 发表于 2018-5-28 13:15
\(\D\lim_{n\to\infty}\frac{1+2+3+4+...+n}{(1+1+1+1+...+1)×n}=\frac{1}{2 ...
万 事 还 得 回 到“1”。
\(\D\lim_{n\to\infty}\frac{1^1+2^1+...+n^1}{n^2}+\lim_{n\to\infty}\frac{1^1+2^1+...+n^1}{(1^0+2^0+...+n^0)×n}=\frac{1}{2}+\frac{1}{2}=1\)
\(\D\lim_{n\to\infty}\frac{1^2+2^2+...+n^2}{n^3}+\lim_{n\to\infty}\frac{1^2+2^2+...+n^2}{(1^1+2^1+...+n^1)×n}=\frac{1}{3}+\frac{2}{3}=1\)
\(\D\lim_{n\to\infty}\frac{1^3+2^3+...+n^3}{n^4}+\lim_{n\to\infty}\frac{1^3+2^3+...+n^3}{(1^2+2^2+...+n^2)×n}=\frac{1}{4}+\frac{3}{4}=1\)
\(\D\lim_{n\to\infty}\frac{1^4+2^4+...+n^4}{n^5}+\lim_{n\to\infty}\frac{1^4+2^4+...+n^4}{(1^3+2^3+...+n^3)×n}=\frac{1}{5}+\frac{4}{5}=1\)
\(\D\lim_{n\to\infty}\frac{1^5+2^5+...+n^5}{n^6}+\lim_{n\to\infty}\frac{1^5+2^5+...+n^5}{(1^4+2^4+...+n^4)×n}=\frac{1}{6}+\frac{5}{6}=1\)
\(\D\lim_{n\to\infty}\frac{1^6+2^6+...+n^6}{n^7}+\lim_{n\to\infty}\frac{1^6+2^6+...+n^6}{(1^5+2^5+...+n^5)×n}=\frac{1}{7}+\frac{6}{7}=1\)
\(\D\lim_{n\to\infty}\frac{1^{e -2}+2^{e -2}+...+n^{e -2}}{n^{e -1}}+\lim_{n\to\infty}\frac{1^{e -2}+2^{e -2}+...+n^{e -2}}{(1^{e -3}+2^{e -3}+...+n^{e -3})×n}=\frac{1}{e -1}+\frac{e -2}{e -1}=1\)
\(\D\lim_{n\to\infty}\frac{1^{e -1}+2^{e -1}+...+n^{e -1}}{n^{e -0}}+\lim_{n\to\infty}\frac{1^{e -1}+2^{e -1}+...+n^{e -1}}{(1^{e -2}+2^{e -2}+...+n^{e -2})×n}=\frac{1}{e -0}+\frac{e -1}{e -0}=1\)
\(\D\lim_{n\to\infty}\frac{1^{e+0}+2^{e+0}+...+n^{e+0}}{n^{e+1}}+\lim_{n\to\infty}\frac{1^{e+0}+2^{e+0}+...+n^{e+0}}{(1^{e -1}+2^{e -1}+...+n^{e -1})×n}=\frac{1}{e+1}+\frac{e -0}{e +1}=1\)
\(\D\lim_{n\to\infty}\frac{1^{e+1}+2^{e+1}+...+n^{e+1}}{n^{e+2}}+\lim_{n\to\infty}\frac{1^{e+1}+2^{e+1}+...+n^{e+1}}{(1^{e+0}+2^{e+0}+...+n^{e=0})×n}=\frac{1}{e+2}+\frac{e+1}{e +2}=1\)
\(\D\lim_{n\to\infty}\frac{1^{e+2}+2^{e+2}+...+n^{e+2}}{n^{e+3}}+\lim_{n\to\infty}\frac{1^{e+2}+2^{e+2}+...+n^{e+2}}{(1^{e+1}+2^{e+1}+...+n^{e+1})×n}=\frac{1}{e+3}+\frac{e+2}{e +3}=1\)
\(\D\lim_{n\to\infty}\frac{1^{e+3}+2^{e+3}+...+n^{e+3}}{n^{e+4}}+\lim_{n\to\infty}\frac{1^{e +3}+2^{e+3}+...+n^{e+3}}{(1^{e+2}+2^{e+2}+...+n^{e+2})×n}=\frac{1}{e+4}+\frac{e+3}{e +4}=1\) 王守恩 发表于 2018-5-29 11:51
万 事 还 得 回 到“1”。
\(\D\lim_{n\to\infty}\frac{1^1+2^1+...+n^1}{n^2}+\lim_{n\to\infty} ...
\(\D\frac{1}{3 !}+\frac{2}{5 !}+\frac{3}{7 !}+\frac{4}{9 !}+\frac{5}{11 !}+...=\frac{1}{2 e}\) 王守恩 发表于 2018-6-2 07:54
\(\D\frac{1}{3 !}+\frac{2}{5 !}+\frac{3}{7 !}+\frac{4}{9 !}+\frac{5}{11 !}+...=\frac{1}{2 e}\)
伟大的 "e"