If $f$ is a multiplicative function, then

(1)

Proof. [Proof of (1)] Expand partial products on right of (1) to obtain by fundamental theorem of arithmetic

where 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 \begin{equation*} \abs{\sum_{n=1}^\infty f(n)-\sum_{P_+(n)<y} f(n)}\leq\sum_{n=y}^\infty \abs{f(n)} \end{equation*}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 $\prod_{p<x} (1+1/(p-1))$
  
  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 \begin{equation*} \zeta(s)=\sum_{n=1}^\infty n^{-s}\qquad\text{for $\Re s>1$}. \end{equation*}Since the series converges absolutely, the Euler product for the zeta function is \begin{equation*} \zeta(s)=\prod_{p}\frac{1}{1-p^{-s}}\qquad\text{for $\Re s>1$}. \end{equation*}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.