cn8888 发表于 2014-4-30 16:19:16

质数的平方的倒数和是多少?

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
Clear["Global`*"];(*Clear all variables*)li =
Table, {k, 1, PrimePi}]; Print@
N, 10];
li = Table, {k, 1, PrimePi}]; Print@
N, 10];
li = Table, {k, 1, PrimePi}]; Print@
N, 10];


0.4483700608

0.4519741916

0.4522263342

那么是多少呢?

cn8888 发表于 2014-4-30 16:21:40

二次方 三次方 四次方呢???????

KeyTo9_Fans 发表于 2014-4-30 17:45:24

这个问题一百多年前就有人提出来了。

参考文献:

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

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/questions/53443/sum-of-the-reciprocal-of-the-primes-squared

wayne 发表于 2014-4-30 19:24:46

Column[{#, N, 50]} & /@ Range]

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