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

if and only if:

$\cmod z \ge 1$

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