You are here
HomeEuler product
Primary tabs
Euler product
If $f$ is a multiplicative function, then
$\sum_{{n=1}}^{\infty}f(n)=\prod_{{p\text{ is prime}}}(1+f(p)+f(p^{2})+\cdots)$  (1) 
provided the sum on the left converges absolutely. The product on the right is called the Euler product for the sum on the left.
Proof of (1).
Expand partial products on right of (1) to obtain by fundamental theorem of arithmetic
$\displaystyle\prod_{{p<y}}(1+f(p)+f(p^{2})+\cdots)$  $\displaystyle=\sum_{{k_{1}}}f(p_{1}^{{k_{1}}})\sum_{{k_{2}}}f(p_{2}^{{k_{2}}})% \cdots\sum_{{k_{t}}}f(p_{t}^{{k_{t}}})$  
$\displaystyle=\sum_{{k_{1},k_{2},\ldots,k_{t}}}f(p_{1}^{{k_{1}}})f(p_{2}^{{k_{% 2}}})\cdots f(p_{t}^{{k_{t}}})$  
$\displaystyle=\sum_{{k_{1},k_{2},\ldots,k_{t}}}f(p_{1}^{{k_{1}}}p_{2}^{{k_{2}}% }\cdots p_{t}^{{k_{t}}})$  
$\displaystyle=\sum_{{P_{+}(n)<y}}f(n)$ 
where $p_{1},p_{2},\ldots,p_{t}$ are all the primes between $1$ and $y$, and $P_{+}(n)$ denotes the largest prime factor of $n$. Since every natural number less than $y$ has no factors exceeding $y$ we have that
$\left\lvert\sum_{{n=1}}^{\infty}f(n)\sum_{{P_{+}(n)<y}}f(n)\right\rvert\leq% \sum_{{n=y}}^{\infty}\left\lvert f(n)\right\rvert$ 
which tends to zero as $y\to\infty$. ∎
Examples

If the function $f$ is defined on prime powers by $f(p^{k})=1/p^{k}$ for all $p<x$ and $f(p^{k})=0$ for all $p\geq x$, then Euler’s product formula allows one to estimate $\prod_{{p<x}}(1+1/(p1))$
$\prod_{{p<x}}\left(1+\frac{1}{p1}\right)=\prod_{{p<x}}\left(1+\frac{1}{p}+% \frac{1}{p^{2}}+\cdots\right)=\sum_{{P_{+}(n)<x}}\frac{1}{n}>\sum_{{n<x}}\frac% {1}{n}>\ln x.$ One of the consequences of this formula is that there are infinitely many primes.

The Riemann zeta function is defined by the means of the series
$\zeta(s)=\sum_{{n=1}}^{\infty}n^{{s}}\qquad\text{for $\Re s>1$}.$ Since the series converges absolutely, the Euler product for the zeta function is
$\zeta(s)=\prod_{{p}}\frac{1}{1p^{{s}}}\qquad\text{for $\Re s>1$}.$ If we set $s=2$, then on the one hand $\zeta(s)=\sum_{n}1/n^{2}$ is $\pi^{2}/6$ (proof is here), an irrational number, and on the other hand $\zeta(2)$ is a product of rational functions of primes. This yields yet another proof of infinitude of primes.
Mathematics Subject Classification
11A05 no label found11A51 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections
Comments
about proving the infinitude of primes from the Euler produc...
You mention how at $s=2$, the fact that the product of all primes yields an irrational number is a proof of the infinitude of primes.
That's true, but there's an easier way to get this result from the Euler product, which doesn't depend on comparatively advanced results like the irrationality of $\Pi$:
The $s=1$ case corresponds to the harmonic series. Proving that this diverges is a standard exercise which students will be familiar with. If there were only finitely many primes, the Euler product would be a nonzero rational number and so the harmonic series would have to converge.