Euler product

If $f$ is a multiplicative function, then

$$\sum _{n=1}^{\mathrm{\infty }}f(n)=\prod _{p\text{ is prime}}(1+f(p)+f(p^2)+\mathrm{\cdots })$$(1)

provided the sum on the left converges absolutely. The producton the right is called the Euler product for the sum on theleft.

Proof of (1).

Expand partial products on right of (1) toobtain by fundamental theorem of arithmetic

$={\displaystyle \sum _{k_1}}f(p_1^{k_1}){\displaystyle \sum _{k_2}}f(p_2^{k_2})\mathrm{\cdots }{\displaystyle \sum _{k_t}}f(p_t^{k_t})$

$={\displaystyle \sum _{k_1,k_2,\mathrm{\dots },k_t}}f(p_1^{k_1})f(p_2^{k_2})\mathrm{\cdots }f(p_t^{k_t})$

$={\displaystyle \sum _{k_1,k_2,\mathrm{\dots },k_t}}f(p_1^{k_1}{p}_2^{k_2}\mathrm{\cdots }{p}_t^{k_t})$

where $p_1,p_2,\mathrm{\dots },p_t$ are all the primes between $1$ and $y$, and $P_{+}(n)$ denotes the largest prime factor of $n$. Sinceevery natural number less than $y$ has no factors exceeding $y$ we havethat

which tends to zero as $y\to \mathrm{\infty }$.∎

Examples

• If the function $f$ is defined on prime powers by$f(p^k)=1/p^k$ for all and $f(p^k)=0$ for all $p\ge x$,then allows one to estimate

  One of the consequences of this formula is that there areinfinitely many primes.

• The Riemann zeta function is defined by the means of theseries

  $$\zeta (s)=\sum _{n=1}^{\mathrm{\infty }}{n}^{-s}\mathit{\hspace{1em}\hspace{1em}}\text{for }\mathrm{\Re }s>1.$$

  Since the series converges absolutely, the Euler product for the zeta function is

  $$\zeta (s)=\prod _p\frac{1}{1-p^{-s}}\mathit{\hspace{1em}\hspace{1em}}\text{for }\mathrm{\Re }s>1.$$

  If we set $s=2$, then on the one hand $\zeta (s)={\sum }_{n}1/n^2$ is${\pi }^2/6$ (proof ishere (http://planetmath.org/ValueOfTheRiemannZetaFunctionAtS2)), anirrational number, and on the other hand $\zeta (2)$ is a product of rational functions of primes. This yields yet another proof of infinitude ofprimes.