How to represent $ \frac{1}{a_n}$ in terms of the generating function of $a_n$? 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.
 A: 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.
A: 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.
