质数的平方的倒数和是多少?

cn8888

2 3 5 7 11............
先平方,
再倒数
再求和
那么和是多少呢?

看结果是收敛的

1. Clear["Global`*"];(*Clear all variables*)li =
   
2. Table[Prime[k], {k, 1, PrimePi[50]}]; Print@
   
3. N[Total@Map[1/#^2 &, li], 10];
   
4. li = Table[Prime[k], {k, 1, PrimePi[500]}]; Print@
   
5. N[Total@Map[1/#^2 &, li], 10];
   
6. li = Table[Prime[k], {k, 1, PrimePi[5000]}]; Print@
   
7. N[Total@Map[1/#^2 &, li], 10];
   
8. 
   
9. 
   
10. 0.4483700608
    
11. 
    
12. 0.4519741916
    
13. 
    
14. 0.4522263342

复制代码

那么是多少呢?

cn8888

二次方 三次方 四次方呢???????

KeyTo9_Fans

这个问题一百多年前就有人提出来了。

参考文献：

[1]C. W. Merrifield, The Sums of the Series of Reciprocals of the Prime Numbers and of Their Powers, Proc. Roy. Soc. London 33 (1881) 4–10. doi:10.1098/rspl.1881.0063. JSTOR 113877

[2]J. W. L. Glaisher, On the Sums of Inverse Powers of the Prime Numbers, Quart. J. Math. 25 (1891) 347–362.

$1$楼要求的是$P(2)=0.4522474200...$，

$2$楼要求的是$P(3)$、$P(4)$、……

其中$P$是“prime zeta function”。

参考链接：

http://mathoverflow.net/question ... -the-primes-squared

wayne

1. Column[{#, N[PrimeZetaP[#], 50]} & /@ Range[2, 10]]

复制代码

1. {2,0.45224742004106549850654336483224793417323134323989}
   
2. {3,0.17476263929944353642311331466570670097541212192615}
   
3. {4,0.076993139764246844942619295933157870162041059714843}
   
4. {5,0.035755017483924257132818242538855711131697276726651}
   
5. {6,0.017070086850636512954133673266059399209585941874544}
   
6. {7,0.0082838328561335925351241387294487230891833288853078}
   
7. {8,0.0040614053665178305605234391426830805229771445120717}
   
8. {9,0.0020044675749624506630735851407831175368229203497386}
   
9. {10,0.00099360357443698021785585070014773941630187254528520}
   

复制代码