求证:\(\frac{S_{1}}{S_{2}}=\frac{4}{5}\)

王守恩

求证:\(\frac{S_{1}}{S_{2}}=\frac{4}{5}\)

\(\displaystyle S_{1}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{5^2}+\frac{1}{7^2}-\frac{1}{8^2}+\frac{1}{10^2}-\frac{1}{11^2}+\cdots\cdots=\sum_{n=1}^{\infty}\bigg(\frac{1}{(3n-2)^2}-\frac{1}{(3n-1)^2}\bigg)\)

\(\displaystyle S_{2}=\frac{1}{1^2}-\frac{1}{5^2}+\frac{1}{7^2}-\frac{1}{11^2}+\frac{1}{13^2}-\frac{1}{17^2}+\frac{1}{19^2}-\frac{1}{23^2}+\cdots\cdots=\sum_{x=1}^{\infty}\bigg(\frac{1}{(6x-5)^2}-\frac{1}{(6x-1)^2}\bigg)\)

\(\displaystyle S_{1}=\sum_{n=1}^{\infty}\bigg(\frac{1}{(3n-2)^2}-\frac{1}{(3n-1)^2}\bigg)=\sum_{x=1}^{\infty}\bigg(\frac{1}{(3(2x-1)-2)^2}-\frac{1}{(3(2x-1)-1)^2}\bigg)+\sum_{x=1}^{\infty}\bigg(\frac{1}{(3(2x)-2)^2}-\frac{1}{(3(2x)-1)^2}\bigg)\)

\(\displaystyle=\sum_{x=1}^{\infty}\bigg(\frac{1}{(6x-5)^2}-\frac{1}{(6x-4)^2}\bigg)+\sum_{x=1}^{\infty}\bigg(\frac{1}{6x-2)^2}-\frac{1}{(6x-1)^2}\bigg)=A+B\)

\(\displaystyle S_{2}=\sum_{x=1}^{\infty}\bigg(\frac{1}{(6x-5)^2}-\frac{1}{(6x-1)^2}\bigg)=\sum_{x=1}^{\infty}\bigg(\frac{1}{(6x-5)^2}-\frac{1}{(6x-4)^2}\bigg)+\sum_{x=1}^{\infty}\bigg(\frac{1}{(6x-4)^2}-\frac{1}{(6x-2)^2}\bigg)+\sum_{x=1}^{\infty}\bigg(\frac{1}{(6x-2)^2}-\frac{1}{(6x-1)^2}\bigg)\)

\(\displaystyle=A+\sum_{x=1}^{\infty}\bigg(\frac{1}{(6x-4)^2}-\frac{1}{(6x-2)^2}\bigg)+B=A+\sum_{x=1}^{\infty}\bigg(\frac{1}{4(3x-2)^2}-\frac{1}{4(3x-1)^2}\bigg)+B=A+\frac{1}{4}*S_{1}+B=A+\frac{1}{4}*(A+B)+B=\frac{5}{4}*(A+B)\)

\(\displaystyle\frac{S_{1}}{S_{2}}=\frac{A+B}{\frac{5}{4}*(A+B)}=\frac{4}{5}\)

或:

\(\displaystyle S_{1}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{5^2}+\frac{1}{7^2}-\frac{1}{8^2}+\frac{1}{10^2}-\frac{1}{11^2}+\frac{1}{13^2}-\frac{1}{14^2}+\frac{1}{16^2}-\frac{1}{17^2}+\frac{1}{19^2}-\frac{1}{20^2}+\frac{1}{22^2}-\frac{1}{23^2}+\cdots\cdots\)

\(\displaystyle S_{2}=\frac{1}{1^2}-\frac{1}{5^2}+\frac{1}{7^2}-\frac{1}{11^2}+\frac{1}{13^2}-\frac{1}{17^2}+\frac{1}{19^2}-\frac{1}{23^2}+\cdots\cdots\)

\(\displaystyle=\frac{1}{1^2}-\frac{1}{2^2}+(\frac{1}{2^2}-\frac{1}{4^2})+\frac{1}{4^2}-\frac{1}{5^2}+\frac{1}{7^2}-\frac{1}{8^2}+(\frac{1}{8^2}-\frac{1}{10^2})+\frac{1}{10^2}-\frac{1}{11^2}+\frac{1}{13^2}-\frac{1}{14^2}+(\frac{1}{14^2}-\frac{1}{16^2})+\frac{1}{16^2}-\frac{1}{17^2}+\frac{1}{19^2}-\frac{1}{20^2}+(\frac{1}{20^2}-\frac{1}{22^2})+\frac{1}{22^2}-\frac{1}{23^2}+\cdots\cdots\)

\(\displaystyle=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{5^2}+\frac{1}{7^2}-\frac{1}{8^2}+\frac{1}{10^2}-\frac{1}{11^2}+\cdots\cdots+\frac{1}{4}*(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{5^2}+\frac{1}{7^2}-\frac{1}{8^2}+\frac{1}{10^2}-\frac{1}{11^2}+\cdots\cdots)\)

王守恩

更一般地

\(\displaystyle\frac{\sum_{n=1}^{\infty}\bigg(\frac{1}{(3n-2)^k}-\frac{1}{(3n-1)^k}\bigg)}{\sum_{x=1}^{\infty}\bigg(\frac{1}{(6x-5)^k}-\frac{1}{(6x-1)^k}\bigg)}=\frac{2^k}{2^k+1}\)

王守恩

来个复杂一点的。

\(\displaystyle\frac{\sum_{x=1}^{\infty}\bigg(\frac{1}{(5x-4)^2}+\frac{1}{(5x-1)^2}-\frac{1}{(5x-2)^2}-\frac{1}{(5x-3)^2}\bigg)}{\sum_{x=1}^{\infty}\bigg(\frac{1}{(10x-9)^2}+\frac{1}{(10x-1)^2}-\frac{1}{(10x-3)^2}-\frac{1}{(10x-7)^2}\bigg)}=\frac{4}{5}\)

王守恩

来个纸老虎。

\(\displaystyle\frac{\sum_{x=1}^{\infty}\bigg(\frac{1}{(11x-10)^2}+\frac{1}{(11x-8)^2}+\frac{1}{(11x-7)^2}+\frac{1}{(11x-6)^2}+\frac{1}{(11x-2)^2}-\frac{1}{(11x-9)^2}-\frac{1}{(11x-5)^2}-\frac{1}{(11x-4)^2}-\frac{1}{(11x-3)^2}-\frac{1}{(11x-1)^2}\bigg)}{\sum_{x=1}^{\infty}\bigg(\frac{1}{(22x-21)^2}+\frac{1}{(22x-19)^2}+\frac{1}{(22x-17)^2}+\frac{1}{(22x-13)^2}+\frac{1}{(22x-7)^2}-\frac{1}{(22x-15)^2}-\frac{1}{(22x-9)^2}-\frac{1}{(22x-5)^2}-\frac{1}{(22x-3)^2}-\frac{1}{(22x-1)^2}\bigg)}=\frac{4}{5}\)

王守恩

\(\frac{\sum_{x=1}^{\infty}\big(\frac{1}{(46x-45)^2}+\frac{1}{(46x-43)^2}+\frac{1}{(46x-37)^2}+\frac{1}{(46x-33)^2}+\frac{1}{(46x-21)^2}+\frac{1}{(46x-19)^2}+\frac{1}{(46x-17)^2}+\frac{1}{(46x-15)^2}+\frac{1}{(46x-11)^2}+\frac{1}{(46x-7)^2}+\frac{1}{(46x-5)^2}-\frac{1}{(46x-41)^2}-\frac{1}{(46x-39)^2}-\frac{1}{(46x-35)^2}-\frac{1}{(46x-31)^2}-\frac{1}{(46x-29)^2}-\frac{1}{(46x-27)^2}-\frac{1}{(46x-25)^2}-\frac{1}{(46x-13)^2}-\frac{1}{(46x-9)^2}-\frac{1}{(46x-3)^2}-\frac{1}{(46x-1)^2}\big)}{\sum_{x=1}^{\infty}\big(\frac{1}{(23x-22)^2}+\frac{1}{(23x-21)^2}+\frac{1}{(23x-20)^2}+\frac{1}{(23x-19)^2}+\frac{1}{(23x-17)^2}+\frac{1}{(23x-15)^2}+\frac{1}{(23x-14)^2}+\frac{1}{(23x-11)^2}+\frac{1}{(23x-10)^2}+\frac{1}{(23x-7)^2}+\frac{1}{(23x-5)^2}-\frac{1}{(23x-18)^2}-\frac{1}{(23x-16)^2}-\frac{1}{(23x-13)^2}-\frac{1}{(23x-12)^2}-\frac{1}{(23x-9)^2}-\frac{1}{(23x-8)^2}-\frac{1}{(23x-6)^2}-\frac{1}{(23x-4)^2}-\frac{1}{(23x-3)^2}-\frac{1}{(23x-2)^2}-\frac{1}{(23x-1)^2}\big)}=\frac{3}{4}\)

王守恩

小结。——也就是两串数——数字串A,  数字串B。

数字串A——Select[Prime[Range[120]], MemberQ[{1, 7}, Mod[#, 8]] &]
{7, 17, 23, 31, 41, 47, 71, 73, 79, 89, 97, 103, 113, 127, 137, 151, 167, 191, 193, 199, 223, 233, 239, 241, 257, 263, 271, 281, 311, 313, 337, 353, 359, 367, 383, 401, 409, 431, 433, 439, 449, 457, 463, 479, 487, 503, 521, 569, 577, 593, 599}
代码A——p = 599; b = 9; R = Select[Range[p - 1], JacobiSymbol[#, p] == 1 &]; T = Complement[Range[p - 1], R]; U = # + p Boole[EvenQ[#]] & /@ R; V = # + p Boole[EvenQ[#]] & /@ T;
N[Sum[Total[1/(2 p n - U)^b] - Total[1/(2 p n - V)^b], {n, 10^2}]/Sum[Total[1/(p n - R)^b] - Total[1/(p n - T)^b], {n, 10^2}], 20] —— p = 599; b = 9; ——可以改。
答案A——0.998046875000000000。
答案A——Table[N[(2^b - 1)/2^b, 20], {b, 9}]
{0.500000000000000000, 0.750000000000000000, 0.875000000000000000, 0.937500000000000000, 0.968750000000000000, 0.984375000000000000, 0.992187500000000000, 0.996093750000000000, 0.998046875000000000}

数字串B——Select[Prime[Range[50]], MemberQ[{3, 5}, Mod[#, 8]] &]
{3, 5, 11, 13, 19, 29, 37, 43, 53, 59, 61, 67, 83, 101, 107, 109, 131, 139, 149, 157, 163, 173, 179, 181, 197, 211, 227, 229, 251, 269, 277, 283, 293, 307, 317, 331, 347, 349, 373, 379, 389, 397, 419, 421, 443, 461, 467, 491, 499, 509, 523, 541}
代码B——p = 499; b = 9; R = Select[Range[p - 1], JacobiSymbol[#, p] == 1 &]; T = Complement[Range[p - 1], R]; U = # + p Boole[EvenQ[#]] & /@ R; V = # + p Boole[EvenQ[#]] & /@ T;
N[Sum[Total[1/(2 p n - U)^b] - Total[1/(2 p n - V)^b], {n, 10^2}]/Sum[Total[1/(p n - R)^b] - Total[1/(p n - T)^b], {n, 10^2}], 20] —— p = 499; b = 9; ——可以改。
答案B——1.001953125000000000。
答案B——Table[N[(2^b + 1)/2^b, 20], {b, 9}]
{1.500000000000000000, 1.250000000000000000, 1.125000000000000000, 1.062500000000000000, 1.031250000000000000, 1.015625000000000000, 1.007812500000000000, 1.003906250000000000, 1.001953125000000000}

答案A + 答案B = 0.99804687500000000000 + 1.0019531250000000000 = 2。——对同一个 b 来说。