Difference between revisions of "Riemann zeta"

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$$\zeta(z) = \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{n^z}.$$
 
$$\zeta(z) = \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{n^z}.$$
  
==Convergence==
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==Properties==
 
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<strong>Proposition:</strong> If $\mathrm{Re} \hspace{2pt} z > 1$, then the series defining $\zeta(z)$ converges.
 
<strong>Proposition:</strong> If $\mathrm{Re} \hspace{2pt} z > 1$, then the series defining $\zeta(z)$ converges.
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<strong>Proof:</strong> █
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<strong>Proposition (Euler Product):</strong> $\zeta(z)=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^z} = \displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}}$
 
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<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █
 
<strong>Proof:</strong> █

Revision as of 00:45, 3 August 2014

Consider the function $\zeta$ defined by the series $$\zeta(z) = \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{n^z}.$$

Properties

Proposition: If $\mathrm{Re} \hspace{2pt} z > 1$, then the series defining $\zeta(z)$ converges.

Proof:

Proposition (Euler Product): $\zeta(z)=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^z} = \displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}}$

Proof:

External links