Divergence of the sum of the reciprocals of the primes

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Discussion Overview

The discussion revolves around the divergence of the sum of the reciprocals of the primes, specifically exploring the theorems and mathematical principles that support this divergence. Participants reference calculus theorems related to infinite products and sums, and discuss their implications in the context of prime numbers.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about the theorem used in a specific proof regarding the divergence of the sum of the reciprocals of the primes, suggesting a connection to an elementary theorem of calculus about products converging to nonzero values.
  • Another participant proposes that the convergence of an infinite product is related to the convergence of the sum of logarithms of its terms, indicating that if the sum of the terms converges, then the product converges as well.
  • There is a mention of a potential issue or confusion regarding a negative sign in the context of the convergence discussion, prompting a request for clarification.
  • A suggestion is made to refer to a specific theorem and a book for further understanding of the topic, indicating that there are additional resources available for exploration.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and clarity regarding the theorems discussed, with some confusion noted about specific aspects. No consensus is reached on the interpretation of the mathematical principles involved.

Contextual Notes

There are references to specific mathematical theorems and resources, but the discussion does not resolve the confusion regarding the negative sign mentioned by one participant. The implications of the theorems discussed remain open to interpretation.

Karamata
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Karamata said:
Hi, can you tell me which theorem they have used here: http://everything2.com/title/proof+that+the+sum+of+the+reciprocals+of+the+primes+diverges

i'm thinking on part: Well, there's an elementary theorem of calculus that a product (1-a1)...(1-ak)... with ak->0 converges to a nonzero value iff the sum a1+...+ak+... converges.

Sorry for bad english :redface:



They are using...that theorem. If you want to see why is true take into consideration everything's positive

here and apply logarithms. Of course, you could also google "infinite products"...:>)

DonAntonio
 
Well, I can say:

Let a_n>0, ~n\in \bf{N}. We can say \displaystyle\prod_{n=1}^{+\infty}(1+a_n) converge iff converge \displaystyle\sum_{n=0}^{+\infty}\log(1+a_n). Also, \displaystyle\sum_{n=0}^{+\infty}\log(1+a_n) converge iff converge \displaystyle\sum_{n=0}^{+\infty}a_n, because \displaystyle\lim_{n\to +\infty}\frac{\log(1+a_n)}{a_n} = 1.

So, we have: \displaystyle\prod_{n=1}^{+\infty}(1+a_n) converge iff converge \displaystyle\sum_{n=0}^{+\infty}a_n.

BUT, here, I have this minus...?!?:rolleyes:
 
Karamata said:
Well, I can say:

Let a_n>0, ~n\in \bf{N}. We can say \displaystyle\prod_{n=1}^{+\infty}(1+a_n) converge iff converge \displaystyle\sum_{n=0}^{+\infty}\log(1+a_n). Also, \displaystyle\sum_{n=0}^{+\infty}\log(1+a_n) converge iff converge \displaystyle\sum_{n=0}^{+\infty}a_n, because \displaystyle\lim_{n\to +\infty}\frac{\log(1+a_n)}{a_n} = 1.

So, we have: \displaystyle\prod_{n=1}^{+\infty}(1+a_n) converge iff converge \displaystyle\sum_{n=0}^{+\infty}a_n.

BUT, here, I have this minus...?!?:rolleyes:



Don't worry about that: check theorem 1 here http://tinyurl.com/7cu4k4p , which sends you to check the nice book by

Knopp "Infinite sequences and series" (see chapter 3, section 7 there)

DonAntonio
 
OK, it seems well.

Thanks.
 

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