# Euler product formula

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

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

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

 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{s}}\,=\,\zeta(s).$ (2)
 $\frac{1}{1-\frac{1}{p_{1}^{s}}}\,=\,1+\frac{1}{p_{1}^{s}}+\frac{1}{p_{1}^{2s}}% +\ldots\,=\,\sum_{\nu_{1}=0}^{\infty}\frac{1}{p_{1}^{\nu_{1}s}},$
 $\frac{1}{1-\frac{1}{p_{2}^{s}}}\,=\,1+\frac{1}{p_{2}^{s}}+\frac{1}{p_{2}^{2s}}% +\ldots\,=\,\sum_{\nu_{2}=0}^{\infty}\frac{1}{p_{2}^{\nu_{2}s}}.$

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

 $\frac{1}{1-\frac{1}{p_{1}^{s}}}\cdot\frac{1}{1-\frac{1}{p_{2}^{s}}}\;=\;\sum_{% \nu_{1},\nu_{2}=0}^{\infty}\frac{1}{p_{1}^{\nu_{1}s}}\cdot\frac{1}{p_{2}^{\nu_% {2}s}}\;=\;\sum_{\nu_{1},\nu_{2}=0}^{\infty}\frac{1}{\left(p_{1}^{\nu_{1}}p_{2% }^{\nu_{2}}\right)^{s}}$

where $\nu_{1}$ and $\nu_{2}$ independently on each other run all nonnegative integers.  This equation can be generalised by induction to

 $\displaystyle\prod_{\nu=1}^{k}\frac{1}{1-\frac{1}{p_{\nu}^{s}}}\;=\;\sum_{\nu_% {1},\nu_{2},\ldots,\nu_{k}=0}^{\infty}\frac{1}{\left(p_{1}^{\nu_{1}}p_{2}^{\nu% _{2}}\cdots p_{k}^{\nu_{k}}\right)^{s}}$ (3)

for  $s>0$  and for arbitrarily great $k$; the exponents  $\nu_{1},\,\nu_{2},\,\ldots,\,\nu_{k}$  run independently all nonnegative integers.

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

 $\displaystyle\prod_{\nu=1}^{k}\frac{1}{1-\frac{1}{p_{\nu}^{s}}}\;=\;\sum_{(n)}% \frac{1}{n^{s}},$ (4)

where $n$ runs all positive integers not containing greater prime factors  than $p_{k}$.  Then the inequality

 $\displaystyle\sum_{n=1}^{p_{k}}\frac{1}{n^{s}}<\prod_{\nu=1}^{k}\frac{1}{1-% \frac{1}{p_{\nu}^{s}}},$ (5)

holds for every $k$, since all the terms  $1,\,\frac{1}{p_{1}^{s}},\,\ldots,\,\frac{1}{p_{k}^{s}}$  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

 $\displaystyle\sum_{n=1}^{p_{k}}\frac{1}{n^{s}}<\prod_{\nu=1}^{k}\frac{1}{1-% \frac{1}{p_{\nu}^{s}}}<\zeta(s).$ (6)

Letting  $k\to\infty$,  we have  $p_{k}\to\infty$,  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

 $\displaystyle\prod_{p}\frac{1}{1-\frac{1}{p^{s}}}\;=\;\zeta(s)\qquad(s>1).$ (7)

## References

• 1 E. Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset III.2.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1940).
Title Euler product formula  EulerProductFormula 2013-03-22 18:39:38 2013-03-22 18:39:38 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 40A20 msc 11M06 msc 11A51 msc 11A41 RiemannZetaFunction EulerProduct