Divergence of the sum of the reciprocals of the primes

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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