Difference between revisions of "Reciprocal Riemann zeta in terms of Mobius"
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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
- 1953: Harry Bateman: Higher Transcendental Functions Volume III ... (previous) ... (next): $\S 17.1.2 (10)$
