Suppose I have the ordinary generating function of $a_n$,
$$ A(x) = \sum_{i} a_i x^i$$
Then from the above expression, is there a way to write the generating function of $\frac{1}{a_n}$?
If there is a way to do it using exponential generating function or dirchlet generating function, then that too is fine.
My guess is that is not possible to get a closed form.
Take $a_n = (n+1)^2$. Then $\displaystyle A(x) = \sum_{n=0}^\infty a_n x^n =\frac{1+x}{(1-x)^3}$ and $\displaystyle B(x) = \sum_{n=0}^\infty\frac{1}{a_n}x^n = \frac{1}{x}\rm{Li}_2(x)$ where $\rm{Li}_2$ is the polylogarithm of order $2$. If there were a closed form for $B(x)$ in terms of $A(x)$ one would have a closed form for $\rm{Li}_2$, but there is no known closed form for $\rm{Li}_2$.
And this is just an example taking a very simple sequence $a_n$. If less trivial sequences are considered many others series whose values are unknown will be automatically found, which is extremely unlikely.
In general the two have very little to do with each other. When $A(x) = \sum a_n x^n$ and $B(x) = \sum b_n x^n$ are two generating functions we call $\sum a_n b_n x^n$ their Hadamard product. So $\sum a_n^{-1} x^n$ could be called the Hadamard inverse. Hadamard products are hard to calculate in general (the only general result I know about them is that Hadamard products of rational functions are rational and the proof is not very direct) and Hadamard inverses even more so.
For example when $a_n = \frac{1}{n!}$ we have $A(x) = e^x$ but the Hadamard inverse $\sum n! x^n$ has zero radius of convergence. Interesting things can still be said about it but it doesn't have much of anything to do with $e^x$, as far as I know.
As another example which is even a polynomial, when $a_n = {m \choose n}$ we have $A(x) = (1 + x)^m$ but the Hadamard inverse $\sum {m \choose n}^{-1} x^n$ is much stranger. It can be represented using the Beta function integral but this doesn't have much to do with $A(x)$, again as far as I know.
