# Sum of Reciprocals of Powers as Euler Product

## Theorem

Let $\zeta$ be the Riemann zeta function.

Let $s\in \C$ be a complex number with real part $\sigma>1$.

Then $\zeta(s) = \displaystyle\prod_p\frac1{1-p^{-s}}$ where the infinite product runs over the prime numbers.

## Proof 1

From Euler Product:

$\displaystyle \sum_{n \mathop \ge 1} a_n n^{-z} = \prod_p \frac 1 {1 - a_p p^{-z} }$

if and only if $\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z}$ is absolutely convergent.

For all $n \in \Z_{\ge 1}$, let $a_n = 1$.

$\displaystyle \sum_{n \mathop \ge 1} n^{-z}$ is absolutely convergent
$\left\lvert{z}\right\rvert \ge 1$

Then it follows that:

$\displaystyle \sum_{n \mathop \ge 1} \frac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$

## Proof 2

$\dfrac 1 {1 - p^{-z} } = 1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \cdots$
$\displaystyle \sum_{n \mathop \ge 1} n^{-z}$ is absolutely convergent
$\left\lvert{z}\right\rvert \ge 1$

Thus:

 $\displaystyle \sum_p \dfrac 1 {1 - p^{-z} }$ $=$ $\displaystyle \sum_p \left({1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \dfrac 1 {p^{3 z} } + \cdots}\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left({1 + \dfrac 1 {2^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {2^{3 z} } + \cdots}\right)$ $\quad$ $\quad$ $\displaystyle$  $\, \displaystyle \times \,$ $\displaystyle \left({1 + \dfrac 1 {3^z} + \dfrac 1 {3^{2 z} } + \dfrac 1 {3^{3 z} } + \cdots}\right)$ $\quad$ $\quad$ $\displaystyle$  $\, \displaystyle \times \,$ $\displaystyle \left({1 + \dfrac 1 {5^z} + \dfrac 1 {5^{2 z} } + \dfrac 1 {5^{3 z} } + \cdots}\right)$ $\quad$ $\quad$ $\displaystyle$  $\, \displaystyle \times \,$ $\displaystyle \cdots$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 1 + \dfrac 1 {2^z} + \dfrac 1 {3^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {5^z}$ $\quad$ $\quad$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \dfrac 1 {2^z 3^z} + \dfrac 1 {7^z} + \dfrac 1 {2^{3 z} } + \dfrac 1 {3^{2 z} }$ $\quad$ $\quad$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \cdots$ $\quad$ $\quad$

The result follows from the Fundamental Theorem of Arithmetic.

$\blacksquare$