Is it possible to get a 'closed form' for $\sum_{k=0}^{n} a_k b_{n-k}$? This came up when trying to divide series, or rather, express $\frac1{f(x)}$ as a series, knowing that $f(x)$ has a zero of order one at $x=0$, and knowing the Taylor series for $f(x)$  (that is knowing the $b_i$ 's).
I write $$1=\frac1{f(x)}f(x) = (\frac{r}{x}+a_0+a_1x+a_2x^2+...)(b_1x+b_2x^2+b_3x^3+...)$$
And comparing coefficients I get $r= \frac{1}{b_1}$ and more importantly, the equations:
$$0=b_{n+1}r + \sum_{k=0}^{n} a_k b_{n-k}.$$
Is there a closed form for this recursion, i.e. is it possible to extract $a_n = ...$ from this? In the particular case I'm looking at, $b_n = \frac{1}{n!}$.

If you know $f$ has a zero of order $1$ you can write it as $$f(x)=x\left(a_1+a_2x+a_3x^2+\ldots\right),$$
with $a_1\neq0$.
Then $\frac{1}{f}=\frac{1}{x}\frac{1}{a_1+a_2x+a_3x^2+\ldots}$.
To compute the series of $\frac{1}{a_1+a_2x+a_3x^2+\ldots}$ just apply long division of $1$ divided by $a_1+a_2x+a_3x^2+...$. This is an algorithm that allows you to compute, term by term, the series if the quotient.
If we have more information on $f$ other methods could be applied. For $f$ a rational function there is a much better method. 
