Sum of Reciprocals of Powers as Euler Product/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\zeta$ be the Riemann zeta function.

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


Then:

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

where the infinite product runs over the prime numbers.


Proof

From Sum of Infinite Geometric Sequence:

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

From Convergence of P-Series:

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

if and only if:

$\cmod z \gt 1$

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.

$\blacksquare$


This article contains statements that are justified by handwavery.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding precise reasons why such statements hold.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Handwaving}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_of_Reciprocals_of_Powers_as_Euler_Product/Proof_2&oldid=563456"