The following equation is for the reciprocal of the zeta function at $Re(s) > 1$:
$$ \frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}} = s \int_{1}^{\infty} \frac{M(x)}{x^{1+s}}dx, \ M(x) = \sum_{n \leq x} \mu(n) $$
Is there a similar equation for the Hurwitz zeta function $Re(s) > 1$:
$$ \frac{1}{\zeta(s,a)} $$
Best Answer
For $q \ge 1,a \ge 1$ integers $$q^{-s}\zeta(s,a/q) = \sum_{n=0}^\infty (qn+a)^{-s}= \sum_{n=1}^\infty c_n n^{-s}$$ is a Dirichlet series but it has no (Euler-like) product so there is no simple expression for its $\log$ and inverse.
Iff $a = 1$ then $c_1 =1\ne 0$ so that $$\frac{1}{q^{-s} \zeta(s,a/q)}= \frac{1}{1-(-\sum_{n=2}^\infty c_n n^{-s})}=1+\sum_{k=1}^\infty (-\sum_{n=2}^\infty c_n n^{-s})^k= 1+\sum_{k=1}^\infty\sum_{n= 2^k}^\infty b_{k,n} n^{-s}$$ is a Dirichlet series.
For $a \ne 1$ the inverse is a generalized Dirichlet series.
If $a/q \not \in \Bbb{Q}$ then both $\zeta(s,a/q), \frac{1}{\zeta(s,a/q)}$ are generalized Dirichlet series.