Sum of Reciprocals of Powers as Euler Product/Proof 2

Theorem

Let $\zeta$ be the Riemann zeta function.

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

Then:

$\displaystyle \map \zeta s = \prod_{\text {$p$prime} } \frac 1 {1 - p^{-s} }$

where the infinite product runs over the prime numbers.

Proof

$\dfrac 1 {1 - p^{-z} } = 1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \cdots$
$\displaystyle \sum_{n \mathop = 1}^\infty n^{-z}$ is absolutely convergent
$\cmod z \gt 1$

Thus:

 $\ds \sum_p \dfrac 1 {1 - p^{-z} }$ $=$ $\ds \sum_p \paren {1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \dfrac 1 {p^{3 z} } + \cdots}$ $\ds$ $=$ $\ds \paren {1 + \dfrac 1 {2^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {2^{3 z} } + \cdots}$ $\ds$  $\, \ds \times \,$ $\ds \paren {1 + \dfrac 1 {3^z} + \dfrac 1 {3^{2 z} } + \dfrac 1 {3^{3 z} } + \cdots}$ $\ds$  $\, \ds \times \,$ $\ds \paren {1 + \dfrac 1 {5^z} + \dfrac 1 {5^{2 z} } + \dfrac 1 {5^{3 z} } + \cdots}$ $\ds$  $\, \ds \times \,$ $\ds \cdots$ $\ds$ $=$ $\ds 1 + \dfrac 1 {2^z} + \dfrac 1 {3^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {5^z}$ $\ds$  $\, \ds + \,$ $\ds \dfrac 1 {2^z 3^z} + \dfrac 1 {7^z} + \dfrac 1 {2^{3 z} } + \dfrac 1 {3^{2 z} }$ $\ds$  $\, \ds + \,$ $\ds \cdots$

The result follows from the Fundamental Theorem of Arithmetic.

$\blacksquare$