I am attending a seminar on Number Theory and we were proving the prime number theorem.
This is the proof we were woring on:
http://people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2975232/fulltext.pdf
Im at the part were we prove the equality:
$$ \sum_{i=1}^{\infty} \frac{1}{n^s} = \prod_p \frac{1}{1 - p^{-s}} $$
So... maybe its pretty trivial but, I don´t understand where this comes from:
$\sum_{r_{1},r_{2},...}^{\infty}\ (2^{r_2}2^{r_3}...)^{-s}$ = $\prod_p\ (\sum_{r≥0} \ p^{-rs} )$
I get how you get to the first part of the equality (i.e. $\sum_{r_{1},r_{2},...}^{\infty}\ (2^{r_2}2^{r_3}...)^{-s}$), because we are working on an unique factorization domain, and because the Riemann zeta-function converges absolutely. But I cannot seem to grasp how do you get that product of sums.
P.S. I am sory if this question is too trivial, but i searched for this proof and cannot seem o find someone who does it this way.
The proof was offered by Euler-see here. Note that we can do the following: $$\zeta(s) = 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+ \ldots $$
$$\frac{1}{2^s}\zeta(s) = \frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\frac{1}{8^s}+\frac{1}{10^s}+ \ldots $$
Subtracting the second equation from the first we remove all elements that have a factor of 2:
$$\left(1-\frac{1}{2^s}\right)\zeta(s) = 1+\frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{9^s}+\frac{1}{11^s}+\frac{1}{13^s}+ \ldots $$
Repeating for the next term:
$$\frac{1}{3^s}\left(1-\frac{1}{2^s}\right)\zeta(s) = \frac{1}{3^s}+\frac{1}{9^s}+\frac{1}{15^s}+\frac{1}{21^s}+\frac{1}{27^s}+\frac{1}{33^s}+ \ldots $$
Subtracting again we get:
$$\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{2^s}\right)\zeta(s) = 1+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{11^s}+\frac{1}{13^s}+\frac{1}{17^s}+ \ldots$$
where all elements having a factor of $3$ or $2$ removed. Continue, until we eventually get $$ \sum_{i=1}^{\infty} \frac{1}{n^s} \times \prod_{p \text{ is prime}}^{\infty} \left(1 - \frac{1}{p^s}\right)=1 $$ Divide each side by the product to get the desired result.