A simple way to obtain $\prod_{p\in\mathbb{P}}\frac{1}{1-p^{-s}}=\sum_{n=1}^{\infty}\frac{1}{n^s}$.

Let $ p_1<p_2 <\cdots <p_k < \cdots $ the increasing list in set $\mathbb{P}$ of all prime numbers . By sum of infinite geometric series we have $\sum_{k=0}^\infty r^k = \frac{1}{1-r}$, for $0<r<1$. For all $s>1$ and $r=\frac{1}{p_k^{s}}$ we have $$ \begin{array}{cccccc} \dfrac{1}{1-p_{1}^{-s}} & = & 1+\dfrac{1}{(p_1^s)^1}+\dfrac{1}{(p_1^s)^2}+\dfrac{1}{(p_1^s)^3}+ & \!\!\cdots\!\! & +\dfrac{1}{(p_1^{s})^{\alpha_1}}+ & \cdots \\ \dfrac{1}{1-p_{2}^{-s}} & = & 1+\dfrac{1}{(p_2^s)^1}+\dfrac{1}{(p_2^s)^2}+\dfrac{1}{(p_2^s)^3}+ & \!\!\cdots\!\! & +\dfrac{1}{(p_2^s)^{\alpha_2}}+ & \cdots \\ \dfrac{1}{1-p_{3}^{-s}} & = & 1+\dfrac{1}{(p_3^s)^1}+\dfrac{1}{(p_3^s)^2}+\dfrac{1}{(p_3^s)^3}+ & \!\!\cdots\!\! & +\dfrac{1}{(p_3^s)^{\alpha_3}}+ & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots &\vdots \\ \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \dfrac{1}{1-p_{k}^{-s}} & = & 1+\dfrac{1}{(p_k^s)^1}+\dfrac{1}{(p_k^s)^2}+\dfrac{1}{(p_k^s)^3}+ & \!\!\cdots\!\! & +\dfrac{1}{(p_k^s)^{\alpha_k}}+ & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \end{array} $$

And the Fundamental Theorem of Arithmetic tells us that every integer $ n> 1$ can be decomposed uniquely as a product $$ n= p_{i_1}^{\alpha_{i_1}}p_{i_2}^{\alpha_{i_2}}\cdots p_{i_k}^{\alpha_{i_k}} $$ of powers of prime numbers $p_{i_1}< p_{i_2}< \cdots < p_{i_k}$ for integers $\alpha_{i_1},\alpha_{i_2},\ldots,\alpha_{i_k}\geq 1$. Since $ n^s= (p_{i_1}^s)^{\alpha_{i_1}}(p_{i_2}^{s})^{\alpha_{i_2}}\cdots (p_{i_k}^s)^{\alpha_{i_k}}$ and using brute force with I can prove that $$ \prod_{p\in\mathbb{P}}\frac{1}{1-p^{-s}}=\sum_{n=1}^\infty \frac{1}{n^s} $$ But I would like to know if there is a simple and elegant way to achieve this result is up through the above list.

Solution 1:

I absolutely love this result, I literally cannot stop from getting goosebumps and smiling whenever I think about it. It is a proof from probability theory! I learned it in David William's Probability with Martingales, of which it is part of exercise E4.2.

Fix $s>1$ and recall that $\zeta(s) = \sum_{n \in \mathbb{N}} n^{-s}$, so we aim to show that $1/\zeta(s) = \prod_p(1-p^{-s})$ where of course $p$ ranges over the primes.

First, define a probability measure $P$ and an $\mathbb{N}$-valued random variable $X$ such that $P(X=n) = n^{-s}/\zeta(s)$ (for example take $P(\{n\}) = n^{-s}/\zeta(s)$ and $X(\omega)=\omega$). Let $E_k := \{X \text{ is divisible by } k\}$. We claim that the events $(E_p : p \text{ prime})$ are independent. We note that $$ P(E_k) = \sum_{i=1}^\infty P(X=ik) = \sum_{i=1}^\infty \frac{(ik)^{-s}}{\zeta(s)} = k^{-s} \frac{\zeta(s)}{\zeta(s)} = k^{-s}. $$

Then if $p_1,\ldots,p_n$ are distinct primes we have $$\bigcap_{i=1}^n E_{p_i} = E_{\prod_{i=1}^np_i},$$ so that $$ P\left(\bigcap_{i=1}^n E_{p_i}\right) = P(E_{\prod_{i=1}^np_i}) = \left(\prod_{i=1}^n p_i \right)^{-s} = \prod_{i=1}^n p_i^{-s} = \prod_{i=1}^n P(E_{p_i}) $$ so our independence claim is proved. Then we note that $1$ is the unique positive integer which is not a multiple of any prime. Hence $$ \frac{1}{\zeta(s)} = P(X=1) = P\left(\bigcap_p E_p^c\right) = \prod_p(1-P(E_p)) = \prod_p(1-p^{-s}). $$

Solution 2:

Let $s$ for which $\Re(s)>1$ then, for all $p \in \mathbb{P}$ we have $$\sum_{k=1}^{\infty} \frac{1}{p^{ks}}=\left(1-\frac{1}{p^s}\right)^{-1}$$

Now let $A(N)$ be the set of all strictly positive numbers such as all prime divisors are at maximum $N$.

Then $$\prod_{p \in \mathbb{P},\text{ }p<N}\left(1-\frac{1}{p^s}\right)^{-1}=\sum_{n \in A(N)}\frac{1}{n^s}$$

Obviously $\{1,...,N\}\subset A(N)$, then

$$\left|\zeta(s)-\prod_{p \in \mathbb{P},\text{ }p<N}\left(1-\frac{1}{p^s}\right)^{-1}\right|\le \sum_{n>N}\frac{1}{n^{\Re{(s)}}}$$

When $N$ goes to infinity it gives the result.