Dividing an infinite power series by another infinite power series Let's say I have two power series
$\,\mathrm{F}\left(x\right) = \sum_{n = 0}^{\infty}\,a_{n}\,x^{n}$ and
$\,\mathrm{G}\left(x\right) = \sum_{n = 0}^{\infty}\,b_{n}\,x^{n}$.
If I define the function
$\displaystyle{\,\mathrm{H}\left(x\right) =
\frac{\mathrm{F}\left(x\right)}{\mathrm{G}\left(x\right)} =
\frac{\sum_{n = 0}^{\infty}\, a_{n}\,x^{n}}{\sum_{n = 0}^{\infty}\, b_{n}\, x^{n}}}$, is there a general way to expand $\,\mathrm{H}$ such that $\,\mathrm{H}\left(x\right) = \sum_{n=0}^{\infty}\,c_{n}\,x^{n}$ ?.
I guess, what i'm asking is if there is a way to get the first few $c_{n}$ coefficients ?. I'm dealing with a physics problem in which I have two such functions $\,\mathrm{F}$, $\,\mathrm{G}$ and I'd like to get the first few terms in the power series $\,\mathrm{H}$.

Since the multiplication of power series is not that hard we can reduce the task in finding the reciprocal $\frac{1
}{G(x)}$ of a power series
\begin{align*}
G(x)=\sum_{n=0}^\infty b_n x^n
\end{align*}
provided $b_0\ne 0$.

According to H.W. Gould's Combinatorial identities, vol. 4 formula (2.27) the following is valid: Let $b_0\ne 0$, then with
\begin{align*}
\frac{1}{G(x)}=\frac{1}{\sum_{n=0}^\infty b_n x^n}=\sum_{n=0}^\infty B_n x^n
\end{align*}
we obtain
\begin{align*}
B_0&=\frac{1}{b_0}\\
B_n&=\frac{1}{b_0^nn!}\left|
\begin{array}{ccccc}
0&nb_1&nb_2&\cdots&nb_n\\
0&(n-1)b_0&(n-1)b_1&\cdots&(n-1)b_{n-1}\\
0&0&(n-2)b_0&\cdots&(n-2)b_{n-2}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&0&0&\cdots&1\\
\end{array}\tag{1}
\right|
\end{align*}
The right-hand side of (1) is the determinant of an $(n\times n)$-matrix.

