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

Let $p_1 <p_2 <\ldots <p_k <\ldots$ the increasing list in set $\mathbb{P}$ of all prime numbers . By Infinite geometric series $\sum_{k=0}^\infty r^k = \frac{1}{1-r}$ for all $s>1$ and $r=\frac{1}{p_k^{-s}}$, \begin{align*} \frac{1}{1-p_1^{-s}}= & 1+\frac{1}{p_1^{1s}}+\ldots +\frac{1}{p_1^{\alpha_1 s}}+\ldots \\ \frac{1}{1-p_2^{-s}}= & 1+\frac{1}{p_2^{1s}}+\ldots +\frac{1}{p_2^{\alpha_2 s}}+\ldots \\ \vdots\qquad & \quad \vdots \qquad\ldots \qquad\vdots \\ \frac{1}{1-p_K^{-s}}= & 1+ \frac{1}{p_K^{1s}}+\ldots +\frac{1}{p_K^{\alpha_K s}}+\ldots \\ \vdots\qquad & \quad \vdots \qquad\ldots \qquad\vdots \end{align*} And the Fundamental Theorem of Arithmetic tells us that every integer $n> 1$ can be decomposed uniquely as a product $$n= p_{i_1}^{\alpha_{i_1}}p_{i_2}^{\alpha_{i_2}}\ldots p_{i_K}^{\alpha_{i_K}}$$ of powers of prime numbers $p_{i_1}< p_{i_2}< \ldots < p_{i_K}$ for integers $\alpha_{i_1},\alpha_{i_2},\ldots,\alpha_{i_K}\geq 1$. Using brute force with three summation, index and subindex 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.

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The powers you have on the geo expansions are wrong. What does "brute force with three summation, index and subindex" mean? What "above list" are you referring to? The fact that $\prod_p (\sum_k p^{-ks})=\sum_n n^{-s}$ is more or less "obvious" via FTA, and thus the Euler product factorization should present itself as elegant and self-evident in this setting. Though there is less elegant bookkeeping to do in terms of rearranging limits if we desire a rigorous proof. Thus it's not clear to me what your question is exactly. –  anon Jun 24 '13 at 0:18
Elias, you're still missing the leading $1$ for the geometric series. Obviously, those terms are essential when no power of $p_i$ appears in $n$. –  Ted Shifrin Jun 24 '13 at 3:40
@anon Many results intuitive has a certain charm to be difficult to prove rigorously. And this is one of them. –  MathOverview Jul 3 '13 at 15:37

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 $$P(\cap_{i=1}^n E_{p_i}) = P(E_{\prod_{i=1}^np_i}) = \left(\prod_{i=1}^n p_i \right)^{-s} = \prod_{i=1}^n p_i^{-s} = \prod_{i=1}^n P(E_{p_i})$$ 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 $$\frac{1}{\zeta(s)} = P(X=1) = P(\cap_{p} E_p^c) = \prod_p(1-P(E_p)) = \prod_p(1-p^{-s}).$$

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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