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The proof of the Euler Product Formula

(Written by AI)

Start with the full sum

ζ(s)=1+12s+13s+14s+15s+…\displaystyle \zeta(s) = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \frac{1}{5^s} + \dots

The sifting

Multiply the entire equation by 12s\displaystyle \frac{1}{2^s}:

12sζ(s)=12s+14s+16s+…\displaystyle \frac{1}{2^s}\zeta(s) = \frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{6^s} + \dots

Subtract this new equation from the original. This removes every term that is a multiple of 2:

(1−12s)ζ(s)=1+13s+15s+17s+…\displaystyle \left(1 - \frac{1}{2^s}\right)\zeta(s) = 1 + \frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{7^s} + \dots

Repeat for every prime

Repeat the process for the next available number, 3:

(1−13s)(1−12s)ζ(s)=1+15s+17s+111s+…\displaystyle \left(1 - \frac{1}{3^s}\right)\left(1 - \frac{1}{2^s}\right)\zeta(s) = 1 + \frac{1}{5^s} + \frac{1}{7^s} + \frac{1}{11^s} + \dots

Now all multiples of 2 and 3 are gone.

If you keep doing this for every prime number (5,7,11,…5, 7, 11, \dots), every term on the right-hand side disappears except for 1.

…(1−15s)(1−13s)(1−12s)ζ(s)=1\displaystyle \dots \left(1 - \frac{1}{5^s}\right)\left(1 - \frac{1}{3^s}\right)\left(1 - \frac{1}{2^s}\right)\zeta(s) = 1

The result

To isolate ζ(s)\zeta(s), move all those prime factors to the other side:

ζ(s)=1(1−2−s)⋅1(1−3−s)⋅1(1−5−s)…\displaystyle \zeta(s) = \frac{1}{(1-2^{-s})} \cdot \frac{1}{(1-3^{-s})} \cdot \frac{1}{(1-5^{-s})} \dots