Reciprocal of Riemann zeta as a sum of Möbius function for Re(z) greater than 1

From specialfunctionswiki
Revision as of 01:34, 22 June 2016 by Tom (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

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