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:
- $\ds \map \zeta s = \prod_{\text {$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
- $\cmod z \gt 1$
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.
$\blacksquare$
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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