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!