Reciprocal of Riemann Zeta Function

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Theorem

For $\Re \left({z}\right) > 1$:

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

where:

$\zeta$ is the Riemann zeta function
$\mu$ is the Möbius function.


Proof

By definition of the Riemann zeta function:

\(\displaystyle \frac 1 {\zeta \left({z}\right)}\) \(=\) \(\displaystyle \prod_{p \text{ prime} } \left({1 - p^{-z} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({1 - \frac 1 {2^z} }\right) \left({1 - \frac 1 {3^z} }\right) \left({1 - \frac 1 {5^z} }\right) \left({1 - \frac 1 {7^z} }\right) \left({1 - \frac 1 {11^z} }\right) \cdots\)


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.

$\blacksquare$