Euler product formula

Theorem (Euler).  If  $s>1$,  the infinite product

${\displaystyle \prod _p}{\displaystyle \frac{1}{1-\frac{1}{p^s}}}$(1)

where $p$ runs the positive rational primes, converges to the sum of the over-harmonic series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{n^s}}=\zeta (s).$(2)

Proof.  Denote the sequence of prime numbers by   For any  $s>0$,  we can form convergent geometric series

$$\frac{1}{1-\frac{1}{p_1^s}}=\mathrm{\hspace{0.17em}1}+\frac{1}{p_1^s}+\frac{1}{p_1^{2s}}+\mathrm{\dots }=\sum _{{\nu }_1=0}^{\mathrm{\infty }}\frac{1}{p_1^{{\nu }_{1}s}},$$

$$\frac{1}{1-\frac{1}{p_2^s}}=\mathrm{\hspace{0.17em}1}+\frac{1}{p_2^s}+\frac{1}{p_2^{2s}}+\mathrm{\dots }=\sum _{{\nu }_2=0}^{\mathrm{\infty }}\frac{1}{p_2^{{\nu }_{2}s}}.$$

Since these series are absolutely convergent, their product (see multiplication of series) may be written as

$$\frac{1}{1-\frac{1}{p_1^s}}\cdot \frac{1}{1-\frac{1}{p_2^s}}=\sum _{{\nu }_1,{\nu }_2=0}^{\mathrm{\infty }}\frac{1}{p_1^{{\nu }_{1}s}}\cdot \frac{1}{p_2^{{\nu }_{2}s}}=\sum _{{\nu }_1,{\nu }_2=0}^{\mathrm{\infty }}\frac{1}{{\left(p_1^{{\nu }_1}{p}_2^{{\nu }_2}\right)}^s}$$

where ${\nu }_1$ and ${\nu }_2$ independently on each other run all nonnegative integers.  This equation can be generalised by induction to

${\displaystyle \prod _{\nu =1}^k}{\displaystyle \frac{1}{1-\frac{1}{p_{\nu }^s}}}={\displaystyle \sum _{{\nu }_1,{\nu }_2,\mathrm{\dots },{\nu }_k=0}^{\mathrm{\infty }}}{\displaystyle \frac{1}{{\left(p_1^{{\nu }_1}{p}_2^{{\nu }_2}\mathrm{\cdots }{p}_k^{{\nu }_k}\right)}^s}}$(3)

for  $s>0$  and for arbitrarily great $k$; the exponents  ${\nu }_1,{\nu }_2,\mathrm{\dots },{\nu }_k$  run independently all nonnegative integers.

Because the prime factorization of positive integers is unique (http://planetmath.org/FundamentalTheoremOfArithmetic), we can rewrite (3) as

${\displaystyle \prod _{\nu =1}^k}{\displaystyle \frac{1}{1-\frac{1}{p_{\nu }^s}}}={\displaystyle \sum _{(n)}}{\displaystyle \frac{1}{n^s}},$(4)

where $n$ runs all positive integers not containing greater prime factors than $p_k$.  Then the inequality

(5)

holds for every $k$, since all the terms  $1,\frac{1}{p_1^s},\mathrm{\dots },\frac{1}{p_k^s}$  are in the series of the right hand side of (4).  On the other hand, this series contains only a part of the terms of (2).  Thus, for  $s>1$,  the product (3) is less than the sum $\zeta (s)$ of the series (2), and consequently

(6)

Letting  $k\to \mathrm{\infty }$,  we have  $p_k\to \mathrm{\infty }$,  and the sum on the left hand side of (6) tends to the limit $\zeta (s)$, therefore also tends the product (3).  Hence we get the limit equation

${\displaystyle \prod _p}{\displaystyle \frac{1}{1-\frac{1}{p^s}}}=\zeta (s)\mathit{\hspace{1em}\hspace{1em}}(s>1).$(7)