Difference between revisions of "Reciprocal of Riemann zeta as a sum of Möbius function for Re(z) greater than 1"
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The following formula holds for $\mathrm{Re}(z)>1$: | The following formula holds for $\mathrm{Re}(z)>1$: | ||
$$\dfrac{1}{\zeta(z)} = \displaystyle\sum_{n=1}^{\infty} \dfrac{\mu(n)}{n^z},$$ | $$\dfrac{1}{\zeta(z)} = \displaystyle\sum_{n=1}^{\infty} \dfrac{\mu(n)}{n^z},$$ | ||
| − | where $\zeta$ | + | where $\zeta$ denotes the [[Riemann zeta function]] and $\mu$ is the [[Möbius function]]. |
==Proof== | ==Proof== | ||
==References== | ==References== | ||
| − | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Möbius function|next=}}: 24.3.1 B | + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Möbius function|next=Identity written as a sum of Möbius functions}}: $24.3.1. \mathrm{I}.B.$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 01:34, 22 June 2016
Theorem
The following formula holds for $\mathrm{Re}(z)>1$: $$\dfrac{1}{\zeta(z)} = \displaystyle\sum_{n=1}^{\infty} \dfrac{\mu(n)}{n^z},$$ where $\zeta$ denotes the Riemann zeta function and $\mu$ is the Möbius function.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $24.3.1. \mathrm{I}.B.$
