Reciprocal of Riemann Zeta Function

Theorem

For $\map \Re z > 1$:

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

where:

$\zeta$ is the Riemann zeta function

$\mu$ is the Möbius function.

Proof

By Sum of Reciprocals of Powers as Euler Product:

\(\ds \frac 1 {\map \zeta z}\)

\(=\)

\(\ds \prod_{\text {$p$ prime} } \paren {1 - p^{-z} }\)

\(\ds \)

\(=\)

\(\ds \paren {1 - \frac 1 {2^z} } \paren {1 - \frac 1 {3^z} } \paren {1 - \frac 1 {5^z} } \paren {1 - \frac 1 {7^z} } \paren {1 - \frac 1 {11^z} } \cdots\)

The expansion of this product will be:

$\ds 1 + \sum_{\text {$n$ prime} } \paren {\frac{-1} {n^z} } + \sum_{n \mathop = p_1 p_2} \paren {\frac {-1} {p_1^z} \frac {-1} {p_2^z} } + \sum_{n \mathop = p_1 p_2 p_3} \paren {\frac {-1} {p_1^z} \frac {-1} {p_2^z} \frac {-1}{p_3^z} } + \cdots$

which is precisely:

$\ds \sum_{n \mathop = 1}^\infty \frac {\map \mu n} {n^z}$

as desired.

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$\blacksquare$