The Euler product formula for the Riemann zeta function reads

${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}}$ 

where the left hand side equals the Riemann zeta function:

${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=1+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+{\frac {1}{4^{s}}}+{\frac {1}{5^{s}}}+\ldots }$ 

and the product on the right hand side extends over all prime numbers p:

${\displaystyle \prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}={\frac {1}{1-2^{-s}}}\cdot {\frac {1}{1-3^{-s}}}\cdot {\frac {1}{1-5^{-s}}}\cdot {\frac {1}{1-7^{-s}}}\cdots {\frac {1}{1-p^{-s}}}\cdots }$ 

This sketch of a proof makes use of simple algebra only. This was the method by which Euler originally discovered the formula. There is a certain sieving property that we can use to our advantage:

${\displaystyle \zeta (s)=1+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+{\frac {1}{4^{s}}}+{\frac {1}{5^{s}}}+\ldots }$ 

${\displaystyle {\frac {1}{2^{s}}}\zeta (s)={\frac {1}{2^{s}}}+{\frac {1}{4^{s}}}+{\frac {1}{6^{s}}}+{\frac {1}{8^{s}}}+{\frac {1}{10^{s}}}+\ldots }$ 

Subtracting the second equation from the first we remove all elements that have a factor of 2:

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

Repeating for the next term:

${\displaystyle {\frac {1}{3^{s}}}\left(1-{\frac {1}{2^{s}}}\right)\zeta (s)={\frac {1}{3^{s}}}+{\frac {1}{9^{s}}}+{\frac {1}{15^{s}}}+{\frac {1}{21^{s}}}+{\frac {1}{27^{s}}}+{\frac {1}{33^{s}}}+\ldots }$ 

Subtracting again we get:

${\displaystyle \left(1-{\frac {1}{3^{s}}}\right)\left(1-{\frac {1}{2^{s}}}\right)\zeta (s)=1+{\frac {1}{5^{s}}}+{\frac {1}{7^{s}}}+{\frac {1}{11^{s}}}+{\frac {1}{13^{s}}}+{\frac {1}{17^{s}}}+\ldots }$ 

where all elements having a factor of 3 or 2 (or both) are removed.

It can be seen that the right side is being sieved. Repeating infinitely for ${\displaystyle {\frac {1}{p^{s}}}}$  where ${\displaystyle p}$  is prime, we get:

${\displaystyle \ldots \left(1-{\frac {1}{11^{s}}}\right)\left(1-{\frac {1}{7^{s}}}\right)\left(1-{\frac {1}{5^{s}}}\right)\left(1-{\frac {1}{3^{s}}}\right)\left(1-{\frac {1}{2^{s}}}\right)\zeta (s)=1}$ 

Dividing both sides by everything but the ζ(s) we obtain:

${\displaystyle \zeta (s)={\frac {1}{\left(1-{\frac {1}{2^{s}}}\right)\left(1-{\frac {1}{3^{s}}}\right)\left(1-{\frac {1}{5^{s}}}\right)\left(1-{\frac {1}{7^{s}}}\right)\left(1-{\frac {1}{11^{s}}}\right)\ldots }}}$ 

This can be written more concisely as an infinite product over all primes p:

${\displaystyle \zeta (s)=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}}$ 

To make this proof rigorous, we need only to observe that when ${\displaystyle \Re (s)>1}$ , the sieved right-hand side approaches 1, which follows immediately from the convergence of the Dirichlet series for ${\displaystyle \zeta (s)}$ .

An interesting result can be found for ζ(1), the harmonic series:

${\displaystyle \ldots \left(1-{\frac {1}{11}}\right)\left(1-{\frac {1}{7}}\right)\left(1-{\frac {1}{5}}\right)\left(1-{\frac {1}{3}}\right)\left(1-{\frac {1}{2}}\right)\zeta (1)=1}$ 

which can also be written as,

${\displaystyle \ldots \left({\frac {10}{11}}\right)\left({\frac {6}{7}}\right)\left({\frac {4}{5}}\right)\left({\frac {2}{3}}\right)\left({\frac {1}{2}}\right)\zeta (1)=1}$ 

which is,

${\displaystyle \left({\frac {\ldots \cdot 10\cdot 6\cdot 4\cdot 2\cdot 1}{\ldots \cdot 11\cdot 7\cdot 5\cdot 3\cdot 2}}\right)\zeta (1)=1}$ 

as, ${\displaystyle \zeta (1)=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\ldots }$ 

thus,

${\displaystyle 1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\ldots ={\frac {2\cdot 3\cdot 5\cdot 7\cdot 11\cdot \ldots }{1\cdot 2\cdot 4\cdot 6\cdot 10\cdot \ldots }}}$ 

While the series ratio test is inconclusive for the left-hand side it may be shown divergent by bounding logarithms. Similarly for the right-hand side the infinite coproduct of reals greater than one does not guarantee divergence, e.g.,

${\displaystyle \lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=e}$ .

Instead, the partial products (whose numerators are primorials) may be bounded, using ln(1+x)≤x, as

${\displaystyle \prod _{k=1}^{n}{\frac {p_{k}}{p_{k}-1}}=e^{-\sum _{k=1}^{n}\ln \left(1-{\frac {1}{p_{k}}}\right)}\geq e^{\sum _{k=1}^{n}{\frac {1}{p_{k}}}},}$ 

so that divergence is clear given the double-logarithmic divergence of the inverse prime series. (Note that Euler's original proof for inverse prime series used just the converse direction to prove the divergence of the inverse prime series based on that of the Euler product and the harmonic series.)

Each factor (for a given prime p) in the product above can be expanded to a geometric series consisting of the reciprocal of p raised to multiples of s, as follows

${\displaystyle {\frac {1}{1-p^{-s}}}=1+{\frac {1}{p^{s}}}+{\frac {1}{p^{2s}}}+{\frac {1}{p^{3s}}}+\ldots +{\frac {1}{p^{ks}}}+\ldots }$ 

When ${\displaystyle \Re (s)>1}$ , this series converges absolutely. Hence we may take a finite number of factors, multiply them together, and rearrange terms. Taking all the primes p up to some prime number limit q, we have

${\displaystyle \left|\zeta (s)-\prod _{p\leq q}\left({\frac {1}{1-p^{-s}}}\right)\right|<\sum _{n=q+1}^{\infty }{\frac {1}{n^{\sigma }}}}$ 

where σ is the real part of s. By the fundamental theorem of arithmetic, the partial product when expanded out gives a sum consisting of those terms n^−s where n is a product of primes less than or equal to q. The inequality results from the fact that therefore only integers larger than q can fail to appear in this expanded out partial product. Since the difference between the partial product and ζ(s) goes to zero when σ > 1, we have convergence in this region.

• Euler product
• Riemann zeta function

• John Derbyshire, Prime Obsession: Bernhard Riemann and The Greatest Unsolved Problem in Mathematics, Joseph Henry Press, 2003, ISBN 978-0-309-08549-6