Difference between revisions of "Euler product for Riemann zeta"

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==Theorem==
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<strong>[[Euler product for Riemann zeta|Theorem]] (Euler Product):</strong> The following formula holds:
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The following formula holds for $\mathrm{Re}(z)>1$:
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$$\zeta(z)=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^z} = \displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}},$$
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$$\zeta(z)=\displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}},$$
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where $\zeta$ is the [[Riemann zeta function]].
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where $\zeta$ denotes [[Riemann zeta]].
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<strong>Proof:</strong> █
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==Proof==
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==References==
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* {{BookReference|The Zeta-Function of Riemann|1930|Edward Charles Titchmarsh|prev=Riemann zeta|next=Series for log(riemann zeta) over primes}}: § Introduction $(2)$
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* {{BookReference|Higher Transcendental Functions Volume III|1953|Harry Bateman|prev=Riemann zeta|next=Reciprocal Riemann zeta in terms of Mobius}}: pg. $170$
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* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Riemann zeta|next=findme}}: $23.2.2$
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 05:00, 16 September 2016

Theorem

The following formula holds for $\mathrm{Re}(z)>1$: $$\zeta(z)=\displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}},$$ where $\zeta$ denotes Riemann zeta.

Proof

References

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