Difference between revisions of "Reciprocal of Riemann zeta as a sum of Möbius function for Re(z) greater than 1"
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(Created page with "==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 $\zeta$ is the Riemann...") |
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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== | ||
Revision as of 01:19, 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): 24.3.1 B
