Sum of Reciprocals of Powers as Euler Product

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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.


Corollary 1

Let $\zeta$ be the Riemann zeta function.

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


Then:

$\ds\prod_{\text {$p$ prime} } \paren {1 + p^{-s} } = \dfrac {\map \zeta s} {\map \zeta {2 s} }$

where the infinite product runs over the prime numbers.


Corollary 2

Let $\zeta$ be the Riemann zeta function.

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


Then:

$\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-s} } {1 - p^{-s} } } = \dfrac {\paren {\map \zeta s}^2} {\map \zeta {2 s} }$


Proof 1

By definition of Euler product:

$\ds \sum_{n \mathop = 1}^\infty a_n n^{-z} = \prod_p \frac 1 {1 - a_p p^{-z} }$

if and only if $\ds \sum_{n \mathop = 1}^\infty a_n n^{-z}$ is absolutely convergent.


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For all $n \in \Z_{\ge 1}$, let $a_n = 1$.

From Convergence of P-Series:

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

if and only if:

$\cmod z > 1$


It then follows that:

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$

$\blacksquare$


Proof 2

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.

$\blacksquare$


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Historical Note

The Riemann $\zeta$ Function presented as the Sum of Reciprocals of Powers as Euler Product was discovered by Leonhard Paul Euler.


Sources

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