Reciprocal of Riemann Zeta Function

Theorem

 * $\displaystyle \frac 1 {\zeta \left({z}\right)} = \sum_{k \mathop = 1}^\infty \frac{\mu \left({k}\right)} {k^z}$

where:
 * $\zeta$ is the Riemann zeta function
 * $\mu$ is the Möbius function.

Proof
By definition of the Riemann zeta function:

The expansion of this product will be:


 * $\displaystyle 1 + \sum_{n \text{ prime}} \left({\frac{-1} {n^z} }\right) + \sum_{n \mathop = p_1 p_2} \left({ \frac{-1}{p_1^z} \frac{-1} {p_2^z} }\right) + \sum_{n \mathop = p_1 p_2 p_3} \left({ \frac {-1} {p_1^z} \frac {-1} {p_2^z} \frac{-1}{p_3^z} }\right) + \cdots$

which is precisely:


 * $\displaystyle \sum_{n \mathop = 1}^\infty \frac{\mu \left({n}\right)} {n^z}$

as desired.