Convergence of products

👨🏻‍🏫 學術台數學

優誌企業研究所

2013-10-31 1:13 pm

Pleaes determine the convergence of the following series:-

圖片參考：http://imgcld.yimg.com/8/n/HA00464689/o/20131031050948.jpg

P be the prime number
P3 be the third prime number (i.e. 5)
P4 be the fourth prime number (i.e. 7)
and so on

更新1:

Does it mean that this is a convergent series with the given number equal to zero?

回答 (4)

CRebecca

2013-10-31 6:21 pm

✔ 最佳答案

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.

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