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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.

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  • $\begingroup$ Let $A_m$ be the positive integers whose prime divisors are $\le m$. Then $$\prod_{p \le m} \frac{1}{1-p^{-s}} =\prod_{p \le m}(1+\sum_{k \ge 1}(p^k)^{-s}) = \sum_{n \in A_m} n^{-s}$$ Hence for $Re(s) > 1$ where everything converges absolutely : $$\zeta(s) = \lim_{m \to \infty} \sum_{n \in A_m} n^{-s}= \lim_{m \to \infty}\prod_{p \le m} \frac{1}{1-p^{-s}}$$ $\endgroup$
    – reuns
    Mar 29, 2017 at 1:25

1 Answer 1

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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.

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