# Euler Product formula for Riemann zeta function proof

In class we introduced Reimann Zeta function

$$\zeta (x)=\sum_{n=1}^{+\infty} \frac{1}{n^x}$$ And we proved its domain was $D=(1,+\infty)$

Now Euler proved that

$$\zeta(x)=\prod_{p\text{ prime}}\frac{1}{1-p^{-x}}$$

By saying $$\zeta(x)=1+\frac{1}{2^x}+\frac{1}{3^x}+... \\ \zeta(x)(\frac{1}{2^x})=\frac{1}{2^x}+\frac{1}{4^x}+... \\ \zeta(x)(1-\frac{1}{2^x})=1+\frac{1}{3^x}+\frac{1}{5^x}+...$$

And so on for every prime number.

However this proof isn't a 'rigorous proof' as my professor says. Why is that and how would one prove this rigorously? Any reference would be helpful. I have seen on wikipedia that to make the proof rigorous we need to observe $\mathfrak{R}(x)>1$ Is that the real part of x or something else?

• it is rigorous if you prove first that both side converge, and that you continue your sieve, finding an induction argument on the $k$th prime. and yes the Euler product is true for any $x \in \mathbb{C}$ such that $Re(x) > 1$, but considering only $x$ real and $> 1$ is what Euler did. – reuns Apr 14 '16 at 21:14
• If you want to make it rigorous, first make sure you know what an infinite product means... – punctured dusk Apr 14 '16 at 21:16
• I guess i do know what infinite product means @barto :) Why would you think otherwise ? – daniels_pa Apr 14 '16 at 21:17
• And regarding the equation the professor wasn't clear that we need to prove it for $x\in \mathbb{C}$ so im going to guess $x\in \mathbb{R}$ and $x>1$. And i'm not quite sure i follow @user1952009 does the induction go by prime numbers ? Do i prove the convergence on both sides for the induction step and basis ? – daniels_pa Apr 14 '16 at 21:20
• The book "Gamma: Exploring Euler's Constant by Julian Havil" -- I am pretty sure that you will read not only the easiest but also the most rigorous proof of the theorem. – user231343 Apr 14 '16 at 21:57

(this is how I'd do it)

consider the formal product and series, then by induction on the $k$th prime : $$\prod_p (1+p^{-x}+p^{-2x}+\ldots) = \sum_n a_n n^{-x}$$

now consider the coefficient $a_1$ : it is clearly $1$, the coefficient $a_2$ : it is clearly $1$, etc. (by the fundamental theorem of arithmetic).

now do the same with $$F_K(x) = \prod_{p \le K} (1+p^{-x}+p^{-2x}+\ldots) = \sum_{n=1}^\infty a_n(K) n^{-x}$$

then if $n = \prod_i p_i^{e_i}$ : $a_n(K) = 1$ if all the $p_i \le K$, otherwise $a_n(K) = 0$.

clearly $F_K(x)$ is well-defined for any $x > 1$ (it is a finite product), and $\lim_{K \to \infty} F_K(x)$ exists too because the logarithm of the infinite product is $-\sum_p \ln(1-p^{-x})$ which is absolutely convergent since $\ln(1-p^{-x}) \sim -p^{-x}$ and that $\sum_p p^{-x} < \sum_{n=1}^\infty n^{-x}$ which is (absolutely) convergent.

finally, $\zeta(x)- F_K(x) = \sum_{n=1}^\infty |a_n(K)-1| n^{-x} > 0$, it is absolutely convergent, it is decreasing in $K$, and it clearly $\to 0$ when $K \to \infty$ since every term $\to 0$.

i.e. :

$$\lim_{K \to \infty} F_K(x) = \prod_p \frac{1}{1-p^{-x}} = \zeta(x) \qquad\qquad (\forall \ x > 1)$$

the proof for every $Re(x) > 1$ is a little more complicated, since we don't have monotone convergence of $\zeta(x)-F_K(x)$ to $0$ but only absolute convergence.

• How do you know that $\prod_{p} \sum_{n} p^{-x}$ can be written as $\sum_{n} \frac{a_n}{n^x}$ and how do you see that those relations for $a_n$ apply since a unique decomposition of every natural number by primes don't seem to imply that (or at least not that i see) – daniels_pa Apr 15 '16 at 10:40
• @daniels_pa : I wrote it, by induction. but it was just to show that formally the equality was true. then I showed how to prove rigorously that the functions of $x$ are the same. and if you don't see how the fundamental theorem of arithmetic applies to $\prod_p (\sum_{\nu = 0}^\infty p^{- \nu x})$ it is not my bad. – reuns Apr 15 '16 at 15:48
• in one word : work more. – reuns Apr 15 '16 at 15:50
• I understand what you wanted but the induction goes like for 2 : $\sum_{n=1}^{+\infty} \frac{a_n}{n^x} = \sum_{n=1}^{+\infty} \frac{1}{2^{nx}}$ that said $a_n$ will be 1 for all $2^k$ where $k\in\mathbb{N}$ From there the induction step was bit problematic for me to get but i did finally understood it. I understand your proof that the functions of x are the same from the start. There is no need to be rude :) Sometimes we are all tired and short with time. – daniels_pa Apr 15 '16 at 19:43
• @daniels_pa : finally, the simplest way you should remember is to say that : $\lim_{N \to \infty} \sum_{n \le N}$ and $\lim_{K \to \infty} \prod_{p \le K}$ are just to different orders for summing $(n^{-x})_{n \in \mathbb{N}^*}$, and since for $Re(x) > 1$ it is an ABSOLUTELY convergent series, hence the order of summation doesn't matter, and the two order of summation give the same result (and hence the same function of $x$ : $\zeta(x)$) – reuns Apr 15 '16 at 20:00

Eulers formula for the Zeta function is, $$\prod_{p \in \mathbb{P}}^{p \le A} \frac1{1- p^{-s}} = \prod_{p \in \mathbb{P}} (\sum_{k=0}^{\infty} p^{-ks})$$

Infinite sums and products may depend on order. Finite do not. Consider the finite products, $$\prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} = \prod_{p \in \mathbb{P}}^{p \le A} \sum_{k=0}^{K} p^{-ks}$$

Product over a sum is the sum over the cartesian products of the products. $$\prod_{a \in A} \sum_{b \in B_a} t_{a,b} = \sum_{c \in \prod_{a \in A} B_a} \prod_{a \in A}t_{a,c_a}$$

where the product of sets $\prod_{a \in A} B_a$ is taken to be a cartesian product.

$$\prod_{p \in \mathbb{P}}^{p \le A} \sum_{k=0}^{K} p^{-ks} = \sum_{v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\}} \prod_{p \in P}^{p \le A}p^{-v_ps} = \sum_{v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\}} (\prod_{p \in P}^{p \le A}p^{-v_p})^{-s}$$

Change the summing variable using, $$\sum_{v \in V } g(f(v)) = \sum_{w \in \{f(v) : v \in V \} } g(w)$$ which is valid only if f is one to one. This is true by Fundamental theorem of arithmetic, as every number has a unique factorization. This gives,

$$\prod_{k=0}^{K} \sum_{p \in \mathbb{P}}^{p \le A} p^{-ks} = \sum_{n \in \{\prod_{p \in P}^{p \le A}p^{-v_p} : v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\} \} } n^{-s}$$

Define Q by, $$Q(A, K) = \{\prod_{p \in P}^{p \le A}p^{-v_p} : v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\} \}$$

No natural number may have a factor greater than itself. Also the highest power it can have for a prime factor is $N = 2^K$, giving $K = \log_2(N)$. Every number from 1..N must have a unique factorization, and that factorization must be constructed by Q.

$$\{1 .. N\} \subset Q(N, log_2(N))$$

The positive natural numbers are given by, $$\lim_{A \to \infty, K \to \infty}Q(A,K) = \mathbb{N+} = \{\prod_{p \in P}p^{-v_p} : v \in \prod_{p \in \mathbb{P}} \{ 0 .. \infty\} \}$$

Then, $$\prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} = \prod_{p \in \mathbb{P}}^{p \le A} (\sum_{k=0}^{K} p^{-ks}) = \sum_{n \in Q(A, K)} n^{-s}$$

Taking limits, $$\lim_{A \to \infty, K \to \infty} \prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} = \prod_{p \in \mathbb{P}} \frac1{1- p^{-s}}$$

$$\lim_{A \to \infty, K \to \infty} \sum_{n \in Q(A, K)} n^{-s} = \sum_{n \in \mathbb{N^+}} n^{-s}$$

The order of summation is not prescribed by the sum. Infinite sums can have different values depending on order. However, $$\sum_{1..n}^{\infty} n^{-s}$$ is absolutely convergent for $\Re(s) > 1$, which guarantees that the sum will converge to the same limit irrespective of order. So $$\prod_{p \in \mathbb{P}} \frac1{1- p^{-s}} = \sum_{n \in \mathbb{N^+}} n^{-s} = \sum_{n = 1}^{\infty} n^{-s} = \zeta(s)$$

An alternative approach compares the limit with, $\zeta(s)$. Consider, $$\prod_{p \in \mathbb{P}} \frac1{1- p^{-s}}- \zeta(s)$$

Then, $$\lim_{A \to \infty, K \to \infty} \prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} - \lim_{N \to \infty} \sum_{n = 1}^{N} n^{-s}$$

Or, $$\lim_{A \to \infty, K \to \infty} \sum_{n \in Q(A, K)} n^{-s} - \lim_{N \to \infty} \sum_{n = 1}^{N} n^{-s}$$

So, $$\lim_{N \to \infty} \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})} n^{-s}$$

$s$ may be complex, $s = u + it$ $$\lim_{N \to \infty} \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})} n^{-u} e^{-it \ln(n)}$$

The magnitude is,

$$\lim_{N \to \infty} | \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})} n^{-s} | <= \lim_{N \to \infty} \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})} n^{-u}$$ $$<= \lim_{N \to \infty}\sum_{n=N+1}^{\infty} n^{-u} = 0$$

As $\sum_{n=0}^{\infty} n^{-u}$ converges absolutely for $u > 1$

So if $s = u + it \wedge u > 1$, $$\prod_{p \in \mathbb{P}} \frac1{1- p^{-s}} = \zeta(s)$$

• Hi, you meant $\displaystyle\prod_{p \in \mathbb{P}} (\sum_{k=0}^\infty p^{-ks})$. For the convergence, the main step is $\displaystyle\zeta(s)-\prod_{p \in \mathbb{P},p \le A} (\sum_{k \ge 0} p^{-ks}) = \sum_{n=1}^\infty n^{-s}1_{\text{Lpf}(n) >A}$ where $\text{Lpf}(n)$ is the largest prime factor – reuns Sep 7 '17 at 1:11
• Isn't that bracketing implied? Sorry I don't understand that convergence argument. We are rearranging the terms on an infinite sum. But the euler product converges. So this rearrangement should not matter. Why do you need this extra step? Please explain. – Peter Driscoll Sep 7 '17 at 1:23
• For $\Re(s) > A$ : $\displaystyle\zeta(s)-\prod_{p \in \mathbb{P}} \frac{1}{1-p^{-s}} = \lim_{A \to \infty} \displaystyle\zeta(s)-\prod_{p \in \mathbb{P},p \le A} \sum_{k \ge 0} p^{-ks} = \lim_{A \to \infty} \sum_{n=1}^\infty n^{-s}1_{\text{Lpf}(n) >A} = 0$. Equivalently, since $\{ n^{-s}\}$ is absolutely summable for $\Re(s)> 1$ then $\lim_{N \to \infty} \sum_{n\le N} n^{-s} = \lim_{A \to \infty} \sum_{\text{Lpf}(n) \le A} n^{-s}$ – reuns Sep 7 '17 at 1:27
• Yes I see that, but why do we need it, seeing as the euler product converges $s > 1$. I thought that if a series converged, you could perform the summation in any order, without changing the value. Just interested to know. – Peter Driscoll Sep 7 '17 at 1:41
• (if a series converges absolutely then the order of summation doesn't matter) This is exactly what I did in the last formula. – reuns Sep 7 '17 at 1:46