Euler's Product form of Riemann Zeta Function

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Theorem

Let $s \in \R: s > 1$.

Then:

$\ds \sum_{k \mathop \in \N_{>0} } \dfrac 1 {k^s} = \prod_{p \mathop \in \Bbb P} \dfrac 1 {1 - 1 / p^s}$

where $\Bbb P$ denotes the set of all prime numbers.


Proof

\(\ds \map \zeta {s }\) \(=\) \(\ds \sum_{k \mathop \in \N_{>0} } \dfrac 1 {k^s}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {1^s} + \dfrac 1 {2^s} + \dfrac 1 {3^s} + \dfrac 1 {4^s} + \dfrac 1 {5^s} + \dfrac 1 {6^s} + \cdots\)

Multiplying both sides above by $\dfrac 1 {2^s}$

\(\ds \dfrac 1 {2^s} \map \zeta {s }\) \(=\) \(\ds \dfrac 1 {2^s} + \dfrac 1 {4^s} + \dfrac 1 {6^s} + \dfrac 1 {8^s} + \dfrac 1 {10^s} + \dfrac 1 {12^s} + \cdots\)

Then subtracting:

\(\ds \map \zeta {s } - \dfrac 1 {2^s} \map \zeta {s }\) \(=\) \(\ds \dfrac 1 {1^s} + \dfrac 1 {3^s} + \dfrac 1 {5^s} + \dfrac 1 {7^s} + \dfrac 1 {9^s} + \dfrac 1 {11^s} + \cdots\)
\(\ds \map \zeta {s } \paren {1 - \dfrac 1 {2^s} }\) \(=\) \(\ds \dfrac 1 {1^s} + \dfrac 1 {3^s} + \dfrac 1 {5^s} + \dfrac 1 {7^s} + \dfrac 1 {9^s} + \dfrac 1 {11^s} + \cdots\)

Multiplying both sides above by $\dfrac 1 {3^s}$

\(\ds \dfrac 1 {3^s} \map \zeta {s } \paren {1 - \dfrac 1 {2^s} }\) \(=\) \(\ds \dfrac 1 {3^s} + \dfrac 1 {9^s} + \dfrac 1 {15^s} + \dfrac 1 {21^s} + \dfrac 1 {27^s} + \dfrac 1 {33^s} + \cdots\)

Then subtracting:

\(\ds \map \zeta {s } \paren {1 - \dfrac 1 {2^s} } \paren {1 - \dfrac 1 {3^s} }\) \(=\) \(\ds \dfrac 1 {1^s} + \dfrac 1 {5^s} + \dfrac 1 {7^s} + \dfrac 1 {11^s} + \dfrac 1 {13^s} + \dfrac 1 {17^s} + \cdots\)


Similar to the Sieve of Eratosthenes, in each subsequent step, we multiply both sides of the equation from the previous step by $\dfrac 1 {p^s}$ and then reduce both sides from the previous step by that amount.
After infinitely many steps, we arrive at:
\(\ds \map \zeta {s } \paren {1 - \dfrac 1 {2^s} } \paren {1 - \dfrac 1 {3^s} } \paren {1 - \dfrac 1 {5^s} } \paren {1 - \dfrac 1 {7^s} } \cdots\) \(=\) \(\ds \dfrac 1 {1^s}\)
\(\ds \map \zeta {s } \prod_{p \mathop \in \Bbb P} \paren {1 - \dfrac 1 {p^s} }\) \(=\) \(\ds \dfrac 1 {1^s}\)
\(\ds \map \zeta {s }\) \(=\) \(\ds \prod_{p \mathop \in \Bbb P} \dfrac 1 {1 - 1 / p^s}\)


$\blacksquare$


Source of Name

This entry was named for Leonhard Paul Euler.


Sources

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