Difference between revisions of "Reciprocal Riemann zeta in terms of Mobius"

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(Created page with "==Theorem== The following formula holds: $$\dfrac{1}{\zeta(z)} = \displaystyle\sum_{k=1}^{\infty} \dfrac{\mu(k)}{k^z},$$ where $\zeta$ denotes the Riemann zeta and $\mu$ d...")
 
 
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==Theorem==
 
==Theorem==
The following formula holds:
+
The following formula holds for $\mathrm{Re}(z)>1$:
 
$$\dfrac{1}{\zeta(z)} = \displaystyle\sum_{k=1}^{\infty} \dfrac{\mu(k)}{k^z},$$
 
$$\dfrac{1}{\zeta(z)} = \displaystyle\sum_{k=1}^{\infty} \dfrac{\mu(k)}{k^z},$$
 
where $\zeta$ denotes the [[Riemann zeta]] and $\mu$ denotes the [[Möbius]] function.
 
where $\zeta$ denotes the [[Riemann zeta]] and $\mu$ denotes the [[Möbius]] function.

Latest revision as of 19:31, 21 June 2017

Theorem

The following formula holds for $\mathrm{Re}(z)>1$: $$\dfrac{1}{\zeta(z)} = \displaystyle\sum_{k=1}^{\infty} \dfrac{\mu(k)}{k^z},$$ where $\zeta$ denotes the Riemann zeta and $\mu$ denotes the Möbius function.

Proof

References