2008年2月13日 星期三

Elementary Proof of "Sum Of Reciprocals Of All Primes Are Divergent"

1 / p1 + 1 / p2 + 1 / p3 + ...也是divergent的。這並不明顯，因為質數比整數疏得多，也就是說 1 / p1 + 1 / p2 + 1 / p3 + ... 比 1 + 1/2 + 1/3 + 1/4 + ... 少了很多項。現在寫出證明於PDF檔內（英文）。

6 則留言:

Kahoo 提到...

Marco_Dick 提到...

koopa koo 提到...

Standard reference for your proof is in Apostal's Introduction to Analytic Number Theory, chapter 1. Another proof, which is longer but less tricky uses Riemann Zeta function, and it's Euler product representation.

Marco_Dick 提到...

Hi Koopa. I read this proof from a piece of A4 paper I printed long time ago. Thanks for pointing out the standard reference.

May you also point out any reference for the proof using Euler Product Representation?

koopa koo 提到...

If I remember correctly, it's in a book called: From Fermat to Minkowski. Lectures on the Theory of Numbers and Its Historical Development by Opolka. You can also find it as an exercise from my notes on ENT.

Singmay 提到...

You can find some more proofs here

http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges

I think the proof that Koopa had mentioned is probably the first one.