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

Let $$p_1 the increasing list in set $$\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. 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.

• Heres a fantastic video/playlist that talks about the Euler product youtube.com/… – Fede Poncio Dec 10 '15 at 4:11

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 $$\frac{1}{\zeta(s)} = P(X=1) = P\left(\bigcap_p E_p^c\right) = \prod_p(1-P(E_p)) = \prod_p(1-p^{-s}).$$

• This is indeed a cool proof! Picky point: how does one justify the independence of infinitely many events (used in the last line) when one has shown only independence of finitely many events? – Greg Martin Jun 28 '13 at 3:48
• @GregMartin Independence of an infinite collection is defined to mean every finite subcollection is independent (which I did show). Then we have $$P(\cap_p E^c_p) = P(\cap_n \cap_{p \leq n} E^c_p) = \lim_n P(\cap_{p \leq n} E^c_p) = \lim_n \prod_{p \leq n} (1-P(E_p)) = \prod_p(1-p^{-s}).$$ Of course $p$ ranges over primes and $n$ ranges over natural numbers (should be clear by context). – nullUser Jun 28 '13 at 4:02
• Note that one never has to worry about convergence of products of probabilities because $a_n := \prod_{i=1}^n P[\cdots]$ is a monotone (decreasing) sequence bounded below by $0$, this is why I left out my previous comment from my original argument. – nullUser Jun 28 '13 at 4:11
• @nullUser Congratulations to replicate here this fantastic race. It displays a fantastic correspondence between primality of numbers and independence of measurable sets. – MathOverview Jul 3 '13 at 15:40
• @ThomasAndrews What part confuses you? Make sure to read my comment above. – nullUser Jul 3 '13 at 19:57

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.