Euler product formula

Question

Why is it that $$\prod_{p\,\le\, y}\sum_{k=0}^\infty p^{-ks} = \sum_{n\,\in\,\{n \,:\, p \,\mid\, n \, \Rightarrow \, p\,\le\,y\}} n^{-s} \text{ ?}$$ I understand that we have a product of finitely many absolute convergent series and therefore rearrangement is allowed, but I do not see how exactly the terms are being rearranged.

Answer 1

Here numbers are rearranged according to their $p$-addic valuations.

One non formal way of seeing this :

Since any integer can be uniquely written in a unique way as a product of primes, by expanding $$(1+2+4+\cdots)(1+3+9+\cdots)\cdots(1+p_i^1+p_i^2+\cdots)\cdots$$

you get all integers once because of the fundamental theorem of arithmetic which states that any integer has a unique prime factorisation.

Similarilly, by expanding

$$(1+{1\over2^s}+{1\over4^s}+\cdots)(1+{1\over3^s}+{1\over9^s}+\cdots)\cdots(1+{1\over p_i^s}+{1\over p_i^{2s}}+\cdots)\cdots$$

You get all numbers of the form $\dfrac 1 {n^s}$.

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One nice (formal) proof relies on (very basic) probability theory and is:

Consider the probability $P_s$ on $\Bbb N ^*$ such that $P_s(\{n\}) = {1\over \zeta(s)n^s}$ for $n\ge1$ where $\zeta(s) = \sum_{n=1}^{+\infty}{1\over n^s}$ is the Riemann zeta function.

What follows is that $$P_s(m\Bbb N^*) = \sum_{n=1}^{+\infty}P_s(mn)=\sum_{n=1}^{+\infty}{1\over \zeta(s)(mn)^s} = {1\over m^s}$$

Now since $m\Bbb N^* \cap m'\Bbb N^* = \text{lcm}(m, m')\Bbb N^*$, we have $$P_s(m\Bbb N^* \cap m'\Bbb N^*) = P_s(m\Bbb N^*)P_s( m'\Bbb N^*) \iff \gcd(m, m')=1$$

Let $(p_n)_{n\in \Bbb N^*}$ be the strictly increasing sequence of all prime numbers. From what we just proved we can deduce that $(p_k \Bbb N^*)_{k \in \Bbb N^*}$ is a familly of mutually independant events (this is the key argument which is equivalent to regrouping by p-adic valuations in a very subtle way!).

Therefore their complementary events are also mutually independant, thus : $$P_s\left(\bigcap_{k=1}^{+\infty} \overline{p_k \Bbb N^*}\right) = \prod_{k=1}^{+\infty}\left(1-{1\over p_k^s}\right)$$

Finally, any integer greater than 2 has at least one prime factor. Therefore $$P_s\left(\bigcap_{k=1}^{+\infty} \overline{p_k \Bbb N^*}\right) = P_s(\{1\}) = {1\over \zeta(s)}$$

Which states that $$\zeta(s) = \prod_{k=1}^{+\infty}\left(1-{1\over p_k^s}\right)^{-1} = \prod_{k=1}^{+\infty}\sum_{i=0}^{+\infty}{1\over p_k^{si}}$$

And proves your identity !

Answer 2

Your product $\displaystyle\prod_{p\,:\,p\,\le\,y}$ specifies $p\le y$, so let's try $y=5$, so that $p\in\{2,3,5\}.$ Your sum $\displaystyle \sum_{n \, : \, p \, \mid \, n,\ p\,\le\,y}$ requires $n$ to be divisible only by those primes, so where looking at integers divisible only by $2$, $3$, or $5$ and by no other primes. \begin{align} & \left( 1 + \frac 1 {2^s} + \frac 1 {4^s} + \frac 1 {8^s} + \cdots \right) \\ {}\times {} & \left( 1 + \frac 1 {3^s} + \frac 1 {9^s} + \frac 1 {27^s} + \cdots \right) \\ {}\times{} & \left( 1 + \frac 1 {5^s} + \frac 1 {25^s} + \frac 1 {125^s} + \cdots \right) \\[10pt] = {} & \text{a sum in which, for example, } \frac 1 {7200^s} \text{ appears because} \\[5pt] & 7200 = 2^5 \times3^3\times5 \text{ has no prime} \\[5pt] & \text{factors besides 2, 3, and 5;} \\[5pt] & \text{thus } \frac 1 {7200^s} = \frac 1 {2^{5s}} \times \frac 1 {3^{2s}} \times \frac 1 {5^s}. \tag a \end{align} Within the expression $\displaystyle \left( 1 + \frac 1 {2^s} + \frac 1 {4^s} + \frac 1 {8^s} + \cdots \right)$ one finds the term $1/2^{5s}$; specifically, it is the sixth term in that sum. Likewise, within the sum $\displaystyle \left( 1 + \frac 1 {3^s} + \frac 1 {9^s} + \frac 1 {27^s} + \cdots \right)$ one finds $1/3^{2s}$; specifically, it is $1/9^s$. And within $\displaystyle \left( 1 + \frac 1 {5^s} + \frac 1 {25^s} + \frac 1 {125^s} + \cdots \right)$ one finds $1/5^s$. Hence the term in the line labeled $(\text{a})$ above appears. And so do all other numbers $1/n^s$ where $n$ is a number divisible only by primes that are $\le y$.

It is necessary to rely on the fact that each number has only one prime factorization: If there were two prime factorizations of $7200$, then we would see $\dfrac 2 {7200^s}$ instead of $\dfrac 1 {7200^s}.$