Inverse Rule for Formal Power Series I am just really starting to get into formal power series and understanding them.  I'm particularly interested in looking at the coefficients generated by the inverse of a formal power series:
$$\left(\sum_{n\ge 0}a_nx^n\right)^{-1}=\sum_{n\ge 0}b_nx^n$$
I first thought that my approach would be looking at 
$$\frac{1}{\sum_{n\ge 0}a_nx^n}$$
But I'm more thinking that since we know that a series is invertible in the ring if $a_0$ is invertible in the ring of coefficients.  Thus, since if we assume it is, and since the unit series is $\{1,0,0,0,....\}$ then we have
$$\left(\sum_{n\ge 0}a_nx^n\right)\left(\sum_{n\ge 0}b_nx^n\right)=1$$
Thus we know that $a_0b_0=1$ and thus $b_0=\frac1{a_0}$.  And for the remaining terms we are just looking at the convolution generated by the Cauchy Product and so
$$0=\sum_{j=0}^ka_jb_{k-j}$$
$$-a_0b_k=\sum_{j=1}^ka_jb_{k-j}$$
$$b_k=\frac{-1}{a_0}\sum_{j=1}^ka_jb_{k-j}$$
And thus we have a recursive definition.
Is there another approach that defines the numbers $b_k$ without recursive means?  Are you forced to only recursive methods when operating on the ring of formal power series to calculate coefficents?

The question is very interesting, and Chris Jones gave a very nice answer (not ugly at all!) that can be interpreted in terms of partitions, as follows.
Without loss of generality, we can assume the zeroth coefficient to be $a_0=1$, so that we are inverting:
$$F = 1 + \displaystyle\sum_{n \geq 1} a_nx^n ,$$ 
and then 
\begin{align*}
F^{-1} &= \displaystyle\sum_{n \geq 0} (-1)^n \Big(\sum_{i \geq 1} a_{i}x^i\Big)^n
\end{align*}
(a consequence of the geometric series expansion: $\frac1{1+w} = \sum_{n\geq 0} (-1)^n w^n$ with $w=\sum_{i \geq 1} a_{i}x^i$.)
Next, use the multinomial theorem to expand the inner sum:
\begin{align*}
F^{-1} &= \displaystyle\sum_{n \geq 0} (-1)^n \Big(\sum_{i \geq 1} a_{i}x^i\Big)^n\\
&= \displaystyle\sum_{n \geq 0} (-1)^n \Big(\sum_{\substack{ \beta_1, \beta_2,\dots,\\\sum_{i} \beta_i = n}} \binom{n}{\beta_1, \beta_2,\dots} \prod_{i \geq 1}  (a_{i}x^i)^{\beta_i} \Big)\\
 &= \displaystyle\sum_{n \geq 0}\sum_{\substack{\beta_1, \beta_2,\dots,\\\sum_{i} \beta_i = n}} \Big((-1)^n\binom{n}{\beta_1,\beta_2,\dots}\prod_{i \geq 1} a_{i}^{\beta_i}\Big) x^{\sum_i i\beta_i}
\end{align*}
where the inner sum is over all natural number sequences $\langle \beta_i\rangle$, and $\binom{n}{\beta_1,\beta_2, \dots}$ is a multinomial coefficient. Grouping terms by exponent on $x$, we have the somewhat-closed form
$$F^{-1} = \sum_{n\geq 0} b_n x^n = \displaystyle\sum_{n \geq 0} \Bigg(\sum_{\substack{\beta_1, \beta_2, \dots\\\sum_{i}i\beta_i= n}} (-1)^{\sum_i \beta_i}\binom{\sum_i \beta_i}{\beta_1, \beta_2, \dots} \prod_{i \geq 1} a_i^{\beta_i}\Bigg) x^n $$ 
Finally, the formula for $F^{-1}$ can be rewritten in terms of partitions of the natural number $n$. 
Indeed, partitions of $n$ (into equal or different non-zero parts) are in bijection with sequences of non-negative integers $(\beta_1,\cdots,\beta_n)$, satisfying $$\sum_{i}i\beta_i= n.$$
This gives the interpretation of each $\beta_i$ as the number of parts of size equal to $i$ (it can, of course, be zero). Moreover $\sum_{i}\beta_i$ becomes now the total number of parts of the partition (usually called the length of the partition).
Then, the formula for the coefficients $b_n$ of $F^{-1}$ is: 
$$b_n = \sum_{\beta=(\beta_i)\in P_n} (-1)^{|\beta|} |\beta| !  \prod_{i=1}^n \frac{a_i^{\beta_i}}{\beta_i !} $$
where $P_n$ denotes the finite set of all partitions of $n$, and $|\beta|$ is the length of the partition $\beta = (\beta_1,\cdots,\beta_n)\in P_n$
