Difference between revisions of "Reciprocal of Riemann zeta as a sum of Möbius function for Re(z) greater than 1"
From specialfunctionswiki
(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...") |
(No difference)
|
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 $\zeta$ is 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
