# Divergence of the sum of the reciprocals of the primes

1. May 6, 2012

2. May 6, 2012

### DonAntonio

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

3. May 6, 2012

### Karamata

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....?!?!?!?:uhh:

4. May 6, 2012

### DonAntonio

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

5. May 7, 2012

### Karamata

OK, it seems well.

Thanks.