# 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).
 $\displaystyle\prod_{p $\displaystyle=\sum_{k_{1}}f(p_{1}^{k_{1}})\sum_{k_{2}}f(p_{2}^{k_{2}})\cdots% \sum_{k_{t}}f(p_{t}^{k_{t}})$ $\displaystyle=\sum_{k_{1},k_{2},\ldots,k_{t}}f(p_{1}^{k_{1}})f(p_{2}^{k_{2}})% \cdots f(p_{t}^{k_{t}})$ $\displaystyle=\sum_{k_{1},k_{2},\ldots,k_{t}}f(p_{1}^{k_{1}}p_{2}^{k_{2}}% \cdots p_{t}^{k_{t}})$ $\displaystyle=\sum_{P_{+}(n)

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)

which tends to zero as $y\to\infty$. ∎

## Examples

Title Euler product EulerProduct 2013-03-22 14:10:58 2013-03-22 14:10:58 bbukh (348) bbukh (348) 8 bbukh (348) Definition msc 11A05 msc 11A51 MultiplicativeFunction RiemannZetaFunction