# Sum of Reciprocals of Powers as Euler Product/Proof 2

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## 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

From Sum of Geometric Progression:

- $\dfrac 1 {1 - p^{-z} } = 1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \cdots$

From Convergence of P-Series:

- $\displaystyle \sum_{n \mathop = 1}^\infty n^{-z}$ is absolutely convergent

- $\cmod z \ge 1$

Thus:

\(\displaystyle \sum_p \dfrac 1 {1 - p^{-z} }\) | \(=\) | \(\displaystyle \sum_p \paren {1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \dfrac 1 {p^{3 z} } + \cdots}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {1 + \dfrac 1 {2^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {2^{3 z} } + \cdots}\) | |||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle \times \, \) | \(\displaystyle \paren {1 + \dfrac 1 {3^z} + \dfrac 1 {3^{2 z} } + \dfrac 1 {3^{3 z} } + \cdots}\) | ||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle \times \, \) | \(\displaystyle \paren {1 + \dfrac 1 {5^z} + \dfrac 1 {5^{2 z} } + \dfrac 1 {5^{3 z} } + \cdots}\) | ||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle \times \, \) | \(\displaystyle \cdots\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1 + \dfrac 1 {2^z} + \dfrac 1 {3^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {5^z}\) | |||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \dfrac 1 {2^z 3^z} + \dfrac 1 {7^z} + \dfrac 1 {2^{3 z} } + \dfrac 1 {3^{2 z} }\) | ||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \cdots\) |

The result follows from the Fundamental Theorem of Arithmetic.

$\blacksquare$

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.19$: The Series $\sum 1/ p_n$ of the Reciprocals of the Primes