# In the Beginning, There Was... Euler's Formula!

I will start this blog the way Bernhard Riemann started his paper: with Euler’s product formula that John Derbyshire called the golden key:

$\zeta(s)=\sum_{n\ge1}n^{-s}=\prod_{p}(1-p^{-s})^{-1}$

This holds for any complex number $$s$$ with $$\Re s > 1$$. If you look up a proof in any modern textbook, you will find a number technical rearrangements that end up in an examination of the absolute convergence on both sides. But actually, the formula is nothing but a fancy way of writing out the Sieve of Eratosthenes. Let’s start by writing out the sum on the left hand side:

$\zeta(s)=1^{-s}+2^{-s}+3^{-s}+4^{-s}+5^{-s}+6^{-s}+7^{-s}+8^{-s}+9^{-s}+10^{-s}+\ldots$

Now, let’s sift out all even terms by multiplying the equation by $$2^{-s}$$

$2^{-s}\zeta(s)=2^{-s}+4^{-s}+6^{-s}+8^{-s}+10^{-s}+12^{-s}+14^{-s}+16^{-s}+18^{-s}+\ldots$

and subtracting the second from the first equation:

$(1-2^{-s})\zeta(s)=1^{-s}+3^{-s}+5^{-s}+7^{-s}+9^{-s}+11^{-s}+13^{-s}+15^{-s}+\ldots$

OK, all even terms are gone, now let’s eliminate all remaining multiples of $$3$$. We multiply by $$3^{-s}$$

$3^{-s}(1-2^{-s})\zeta(s)=3^{-s}+9^{-s}+15^{-s}+21^{-s}+27^{-s}+33^{-s}+39^{-s}+45^{-s}+\ldots$

and subtract again:

$(1-3^{-s})(1-2^{-s})\zeta(s)=1^{-s}+5^{-s}+7^{-s}+9^{-s}+11^{-s}+13^{-s}+\ldots$

I think by now the pattern is clear. We continue by multiplying all the primes $$5^{-s}$$, $$7^{-s}$$, $$11^{-s}$$, …, and continue subtracting from what we’ve got before, and arrive at

$\cdots(1-17^{-s})(1-13^{-s})(1-11^{-s})(1-7^{-s})(1-5^{-s})(1-3^{-s})(1-2^{-s})\zeta(s)=1$

Dividing by the factors on the left hand side, we arrive at our final result:

$\zeta(s)=(1-2^{-s})^{-1}(1-3^{-s})^{-1}(1-5^{-s})^{-1}(1-7^{-s})^{-1}\cdots=\prod_{p}(1-p^{-s})^{-1}$

Magic!