王守恩 发表于 2017-1-17 11:03:18

求和

S=1/4+1/8+1/9+1/16+1/25+1/27+1/32+1/36+1/49+1/64+1/81+1/100+1/121+1/125+1/144+1/169+……………………

zeroieme 发表于 2017-1-17 12:56:46

无穷

mathe 发表于 2017-1-17 23:01:13

显然收敛

wayne 发表于 2017-1-18 00:06:50

@mathe,我总结的规律是可以表达成单个数的幂的形式的所有整数的倒数和。而这个存在多因子的合数的幂的情况,应该大于所有单因子数的幂的倒数和,而质数是无穷多,所以这个是发散的。

lsr314 发表于 2017-1-18 09:21:28

$S<(1/2^2+1/2^3+1/2^4+……)+(1/3^2+1/3^3+1/3^4+……)+(1/4^2+1/4^3+1/4^4+……)+……
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……
=1.$

wayne 发表于 2017-1-18 10:00:40

奥对对,首项应该是从平方开始,所以, $S < \sum_{n=2}^{+\infty}{1/n^2}/{1-1/n}= 1 $

王守恩 发表于 2022-12-30 09:18:56

mathe 发表于 2017-1-17 23:01
显然收敛

谢谢《数学研发论坛》!谢谢各路大侠!这些年进步是很大。

王守恩 发表于 2023-1-2 11:54:57

S=1/4+1/8+1/9+1/16+1/25+1/27+1/32+1/36+1/49+1/64+1/81+1/100+1/121+1/125+1/144+1/169+……………………
=0.874464368404944866694351320597373165935338431924214...

A072102
{8, 7, 4, 4, 6, 4, 3, 6, 8, 4, 0, 4, 9, 4, 4, 8, 6, 6, 6, 9, 4, 3, 5, 1, 3, 2, 0, 5, 9, 7, 3, 7, 3, 1, 6, 5, 9, 3, 5, 3, 3, 8, 4, 3, 1, 9, 2, 4, 2, 1,
4, 5, 7, 7, 6, 2, 5, 7, 8, 8, 2, 5, 3, 5, 0, 9, 3, 7, 0, 0, 6, 4, 1, 2, 9, 7, 2, 3, 6, 7, 6, 5, 9, 9, 3, 3, 2, 2, 6, 1, 7, 8, 5, 7, 5, 8, 0, 1, 6, 2,
8, 7, 7, 0, 6, 3, 4, 1, 9, 3, 6, 2, 5, 5, 9, 0, 5, 3, 0, 1, 0, 3, 8, 6, 3, 2, 8, 5, 5, 3, 6, 0, 8, 7, 5, 3, 1, 1, 8, 7, 2, 9, 5, 6, 8, 5, 4, 3, 1, 0}

RealDigits & /@ Table (1 - Zeta), {k, 2, 10^3}]]]][]


A001597
1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, 196, 216, 225, 243, 256, 289, 324, 343, 361, 400,
441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024, 1089, 1156, 1225, 1296, 1331, 1369, 1444, 1521,
1600, 1681, 1728, 1764, 1849, 1936, 2025, 2048, 2116, 2187, 2197, 2209, 2304, 2401, 2500, 2601, 2704, 2744, 2809, 2916,
3025, 3125, 3136, 3249, 3364, 3375, 3481, 3600, 3721, 3844, 3969, 4096, 4225, 4356, 4489, 4624, 4761, 4900, 4913, ......

Join[{1}, Select, GCD@@FactorInteger[#][]>1&]]       (* Harvey P. Dale, Apr 30 2018 *)

王守恩 发表于 2023-1-9 18:02:30

谢谢 mathe!   谢谢 天山草!

N\(\bigg[\D\sum_{k=2}^{166}\big(1-Zeta\big)*MoebiusMu, 50\bigg]\)

0.87446436840494486669435132059737316593533843192421

166可以达到50位(165还不行),331可以达到100位(330还不行),

0.87446436840494486669435132059737316593533843192421
   45776257882535093700641297236765993322617857580163

王守恩 发表于 2023-1-10 09:46:23

谢谢 mathe!   谢谢 天山草!谢谢 Nicolas2050!

太好了!多年的心事总算了了!给出前20项,大家一起分享!谢谢大家!

s(02)=(1-1.6449340668482264365...)×(-1)=+0.6449340668482264365...
s(03)=(1-1.2020569031595942854...)×(-1)=+0.2020569031595942854...
s(04)=(1-1.0823232337111381915...)×(00)=00
s(05)=(1-1.0369277551433699263...)×(-1)=+0.0369277551433699263...
s(06)=(1-1.0173430619844491397...)×(+1)=-0.0173430619844491397...
s(07)=(1-1.0083492773819228268...)×(-1)=+0.0083492773819228268...
s(08)=(1-1.0040773561979443394...)×(00)=00
s(09)=(1-1.0020083928260822144...)×(00)=00
s(10)=(1-1.0009945751278180853...)×(+1)=-0.0009945751278180853...
s(11)=(1-1.0004941886041194646...)×(-1)=+0.0004941886041194646...
s(12)=(1-1.0002460865533080483...)×(00)=00
s(13)=(1-1.0001227133475784891...)×(-1)=+0.0001227133475784891...
s(14)=(1-1.0000612481350587048...)×(+1)=-0.0000612481350587048...
s(15)=(1-1.0000305882363070205...)×(+1)=-0.0000305882363070205...
s(16)=(1-1.0000152822594086519...)×(00)=00
s(17)=(1-1.0000076371976378998...)×(-1)=+0.0000076371976378998...
s(18)=(1-1.0000038172932649998...)×(00)=00
s(19)=(1-1.0000019082127165539...)×(-1)=+0.0000019082127165539...
s(20)=(1-1.0000009539620338728...)×(00)=00
s(21)=(1-1.0000004769329867878...)×(+1)=-0.0000004769329867878...

s(02)+s(03)+s(04)+ .... +s(19)+s(20)+s(21)=+0.87446449947854614427..
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