Convergence of products

圖片參考：http://imgcld.yimg.com/8/n/HA06881079/o/20131031101950.jpg

Ans: the product converges to zero.

2013-10-31 11:25:43 補充：
corr.
П_(n=1~∞) 1/(1-1/pn)=П_(n=1~∞) (1+ 1/pn+ 1/pn^2+...)=Σ_(n=1~∞) 1/n =∞
so П_(n=1~∞) (1- 1/pn) = 0 = (1-1/2)(1-1/3)П_(n=3~∞), then П_(n=3~∞) = 0
0 < = П_(n=3~∞) (1- 2/pn) < = П_(n=3~∞) (1 - 1/pn) =0
hence, П_(n=3~∞) (1- 2/pn) = 0 conv.

2013-10-31 11:31:06 補充：
П_(n=1~∞) 1/(1-1/pn)=П_(n=1~∞) (1+ 1/pn+ 1/pn^2+...)=Σ_(n=1~∞) 1/n =∞
so П_(n=1~∞) (1- 1/pn) = 0 = (1-1/2)(1-1/3)П_(n=3~∞) (1-1/pn), then П_(n=3~∞) (1-1/pn)= 0
0 < = П_(n=3~∞) (1- 2/pn) < = П_(n=3~∞) (1 - 1/pn) =0
hence, П_(n=3~∞) (1- 2/pn) = 0 conv.

2013-10-31 16:15:48 補充：
Yes, the product series converges to zero.

not necessary to use the nature of prime number

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So the second equality is the key.

http://math.stackexchange.com/questions/99007/rate-of-convergence-of-series-of-squared-prime-reciprocals

http://www.mersenneforum.org/showthread.php?t=13437

http://mathworld.wolfram.com/PrimeZetaFunction.html