Determining the coefficients of the reciprocal of a Dirichlet series Given a Dirichlet series with coefficients
$$ F(s)= \sum_{n=1}^{\infty}\frac{a(n)}{n^{s}},$$
is it then always possible (and how) to obtain the $ b(n) $ coefficients related to its reciprocal
$$ \frac{1}{F(s)}=\sum_{n=1}^{\infty}\frac{b(n)}{n^{s}},$$
can we get $ b(n) $ from $ a(n) $ ??
 A: It is possible if $1/F$ has a representation as a Dirichlet series. To obtain the coefficients, it is useful to know how the product of two Dirichlet series looks. Let hence
$$F(s) = \sum_{n = 1}^{\infty} \frac{a(n)}{n^s}\quad \text{and}\quad G(s) = \sum_{n = 1}^{\infty} \frac{b(n)}{n^s}$$
where both series are absolutely convergent for $\operatorname{Re} s > \sigma_0$. Then we can rearrange and regroup freely for $\operatorname{Re} s > \sigma_0$, and obtain
\begin{align}
F(s)\cdot G(s) &= \Biggl(\sum_{k = 1}^{\infty} \frac{a(k)}{k^s}\Biggr)\cdot \Biggl(\sum_{m = 1}^{\infty} \frac{b(m)}{m^s}\Biggr)\\
&= \sum_{k,m = 1}^{\infty} \frac{a(k)b(m)}{(km)^s}\\
&= \sum_{n = 1}^{\infty} \Biggl(\sum_{k m = n} a(k)b(m)\Biggr)\frac{1}{n^s}\\
&= \sum_{n = 1}^{\infty} \Biggl(\sum_{k \mid n} a(k) b(n/k)\Biggr)\frac{1}{n^s}.
\end{align}
So the product of two Dirichlet series is again a Dirichlet series, and the coefficients $c(n)$ of the product satisfy the relation
$$c(n) = \sum_{k \mid n} a(k)b(n/k),$$
that is, the coefficients of the product are given by the Dirichlet convolution of the coefficients of the factors.
In particular, if $G(s) = \frac{1}{F(s)}$, the product is the constant function $1$, and hence
$$c(n) = \begin{cases}1 &, n = 1\\ 0 &, n > 1.\end{cases}\tag{1}$$
The coefficients of the reciprocal of $F$, if that is representable by a Dirichlet series, are given by the Dirichlet inverse of the coefficients of $F$.
We find that a necessary condition is that $a(1) \neq 0$, and that condition is also sufficient. We can then compute the coefficients of the reciprocal with the recurrence $1$, finding $b(1) = \frac{1}{a(1)}$ and recursively
$$b(n) = -\frac{1}{a(1)} \sum_{\substack{k \mid n\\ k < n}} b(k)a(n/k).$$
If the coefficient sequence $(a(n))$ of $F$ has nice properties - for example if it is multiplicative - one may be able to compute the Dirichlet inverse more efficiently. For multiplicative $(a(n))$, one need only consider prime powers, since the Dirichlet inverse is again multiplicative. The coefficients for arbitrary $n$ are then obtained by multiplying the coefficients for the appropriate prime powers.
A: $$F(s) = a_1+\sum_{n=2}^\infty a_n n^{-s}$$


*

*if $a_1 \ne 0$ : we can consider the Dirichlet series $F(s)/a_1$ so without loss of generality we suppose $a_1 = 1$. $F(s)$ is supposed to be convergent at $s=\sigma+it$ so $a_n = o(n^{\sigma})$ so $F(s)$ is absolutely convergent for $Re(s) > \sigma+1 = \sigma_0$ and when $Re(s) \to \infty$ : $F(s) = 1+\mathcal{O}(2^{\sigma_0-Re(s)})$. hence, there exists $\sigma_1$ such that for $Re(s) > \sigma_1$, $|F(s) - 1| < 1$ and we can write without fear : 
$$G(s) = \frac{1}{F(s)} =\frac{1}{1-(1-F(s))} = \sum_{k=0}^\infty (1-F(s))^k = 1+\sum_{k=1}^\infty \left(-\sum_{n=2}^\infty a_n n^{-s}\right)^k$$ $$=1+\sum_{k=1}^\infty \left(\sum_{n=2}^\infty \ldots n^{-s}\right) = 1+\sum_{n=2}^\infty n^{-s} \sum_{n =d_1 \ldots d_k} (-1)^k \prod_{i=1}^k a_{d_i}= \sum_{n=1}^\infty b_n n^{-s}$$ where $\sum_{n =d_1 \ldots d_k}$ is the sum other all the possible factorizations of $n$ where the order counts and each $d_i \ge 2$. the fact that for $Re(s) > \sigma_1$, $|1-F(s)|<1$ implies that $G(s)$ converges for $Re(s)>\sigma_1$ i.e. $b_n = \mathcal{O}(n^{\sigma_1})$ (not saying $G(s)$ has to diverge for $Re(s) < \sigma_1$, it is only a lower bound of its abscissa of convergence, that's all the point of the prime number theorem and the Riemann hypothesis)

*if $a_1 = 0$ : the same argument as above shows that when $\Re(s) \to \infty$, $F(s) \to 0$ so $1/F(s)$ cannot be a Dirichlet series (nor a Mellin/Laplace transform). but, we can divide $F(s)$ by $(n_0)^{-s}$ where $a_{n_0}$ is the smallest non-zero coefficient of $F(s)$ to get a generalized Dirichlet series from which we get : 
$$\frac{(n_0)^{-s}}{F(s)} = G(s) = \sum_{k=0}^\infty (1-F(s)(n_0)^{-s})^k$$
which is also a generalized Dirichlet series, and to which the same arguments as above apply.
