γ是平常数

elim

其实 1分成相同的两半人们也完不成。但数学可以做这些。老头推翻数学？没门!

王守恩

蛮好玩的！

\((\frac{2}{1}-\frac{2}{1}+\frac{2}{3})+(\frac{2}{5}-\frac{2}{3}+\frac{2}{7})+(\frac{2}{9}-\frac{2}{5}+\frac{2}{11})+(\frac{2}{13}-\frac{2}{7}+\frac{2}{15})+......=\ln(2)\)

\((\frac{1}{1}-\frac{2}{2}+\frac{2}{3})+(\frac{1}{2}-\frac{2}{3}+\frac{2}{5})+(\frac{1}{3}-\frac{2}{4}+\frac{2}{7})\ +(\frac{1}{4}\ -\frac{2}{5}+\frac{2}{9})\ +......=\ln(4)\)

\((\frac{2}{1}-\frac{1}{1}+\frac{2}{3})+(\frac{2}{5}-\frac{1}{2}+\frac{2}{7})+(\frac{2}{9}-\frac{1}{3}+\frac{2}{11})+(\frac{2}{13}-\frac{1}{4}+\frac{2}{15})+......=\ln(8)\)

\((\frac{2}{1}-\frac{2}{2}+\frac{2}{3})+(\frac{2}{3}-\frac{2}{3}+\frac{2}{5})+(\frac{2}{5}-\frac{2}{4}+\frac{2}{7})\ +(\frac{2}{7}\ -\frac{2}{5}+\frac{2}{9})\ +......=\ln(16)\)

王守恩

\(要算\ \ln2(2是可以换的)，可以是这样(可以往后走，可以往下走(3)+(4))。别问(看不懂看前面的楼层)\)
(1)基本:
\(\ln2<\frac{1}{1},\frac{1}{2}+\frac{1}{3},\frac{1}{3}+\frac{1}{4}+\frac{1}{5},\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7},\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{09},\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{09}+\frac{1}{10}+\frac{1}{11},\)
(2)基本:
\(\ln2>\frac{1}{2},\frac{1}{3}+\frac{1}{4},\frac{1}{4}+\frac{1}{5}+\frac{1}{6},\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8},\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10},\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12},\)
(3)=\(\frac{(2同列)+(1同列)}{2}:\)
\(\ln2<\frac{1}{2}+\frac{1}{4},\frac{1}{4}+\frac{1}{3}+\frac{1}{08},\frac{1}{6}+\frac{1}{4}+\frac{1}{5}+\frac{1}{12},\frac{1}{8}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{16},\frac{1}{10}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{20},\)
(4)=\(\frac{(2前列)+(1后列)}{2}:\)
\(\ln2>\frac{1}{2}+\frac{1}{6},\frac{1}{3}+\frac{1}{4}+\frac{1}{10},\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{14},\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{18},\frac{1}{06}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{22},\)

王守恩

\(证:\ \displaystyle\sum_{k=1}^{\infty}\frac{i^{2\lceil(Mod(k,A)-A)/A\rceil}}{k}=\ln(A)\ \ \ \ A=2, 3, 4, 5, 6, 7, 8, 9,...\)

\(证:\ \displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{\lfloor k/A\rfloor-\lfloor(k-1)/A\rfloor}}{k}=\ln(A)\ \ \ \ A=2, 3, 4, 5, 6, 7, 8, 9,...\)

王守恩

\(\frac{1}{2}=\frac{\frac{1}{1+2521}+\frac{1}{2+2522}+\frac{1}{3+2523}+\frac{1}{4+2524}+\cdots+\frac{1}{1260+3780}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{2519}-\frac{1}{2520}}\)

\(\frac{1}{2}=\frac{\frac{1}{1+1681}+\frac{1}{2+1682}+\frac{1}{3+1683}+\frac{1}{4+1684}+\cdots+\frac{1}{1680+3360}}{1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\cdots+\frac{1}{2518}+\frac{1}{2519}-\frac{2}{2520}}\)

\(\frac{1}{2}=\frac{\frac{1}{1+1261}+\frac{1}{2+1262}+\frac{1}{3+1263}+\frac{1}{4+1264}+\cdots+\frac{1}{1890+3150}}{1+\frac{1}{2}+\frac{1}{3}-\frac{3}{4}+\cdots+\frac{1}{2517}+\frac{1}{2518}+\frac{1}{2519}-\frac{3}{2520}}\)

\(\frac{1}{2}=\frac{\frac{1}{1+1009}+\frac{1}{2+1010}+\frac{1}{3+1011}+\frac{1}{4+1012}+\cdots+\frac{1}{2016+3024}}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{4}{5}+\cdots+\frac{1}{2516}+\frac{1}{2517}+\frac{1}{2518}+\frac{1}{2519}-\frac{4}{2520}}\)

\(\frac{1}{2}=\frac{\frac{1}{1+841}+\frac{1}{2+842}+\frac{1}{3+843}+\frac{1}{4+844}+\cdots+\frac{1}{2100+2940}}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{5}{6}+\cdots+\frac{1}{2515}+\frac{1}{2516}+\frac{1}{2517}+\frac{1}{2518}+\frac{1}{2519}-\frac{5}{2520}}\)

\(\frac{1}{2}=\frac{\frac{1}{1+721}+\frac{1}{2+722}+\frac{1}{3+723}+\frac{1}{4+724}+\cdots+\frac{1}{2160+2880}}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{6}{7}+\cdots+\frac{1}{2514}+\frac{1}{2515}+\frac{1}{2516}+\frac{1}{2517}+\frac{1}{2518}+\frac{1}{2519}-\frac{6}{2520}}\)

\(\frac{1}{2}=\frac{\frac{1}{1+631}+\frac{1}{2+632}+\frac{1}{3+633}+\frac{1}{4+634}+\cdots+\frac{1}{2205+2835}}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{7}{8}+\cdots+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{7}{2520}}\)

\(\frac{1}{2}=\frac{\frac{1}{1+561}+\frac{1}{2+562}+\frac{1}{3+563}+\frac{1}{4+564}+\cdots+\frac{1}{2240+2800}}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{8}{9}+\cdots+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{8}{2520}}\)

\(\frac{1}{2}=\frac{\frac{1}{1+501}+\frac{1}{2+502}+\frac{1}{3+503}+\frac{1}{4+504}+\cdots+\frac{1}{2270+2770}}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{9}{10}+\cdots+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{9}{2520}}\)