This is a comment on Fred Hucht's answer which doesn't fit in a comment box: Let $a_0$, $a_1$, $a_2$ ... be any sequence of real numbers with $a_0 = 1$ and let $b_0$, $b_1$, $b_2$ ... be defined by $$\sum b_k x^k = \left( \sum a_k x^k \right)^{-1}.$$ Then $$b_k = (-1)^k \det \begin{bmatrix} a_1 & a_2 & a_3 & \cdots & a_k & a_{k+1} \\ 1 & a_1 & a_2 & \cdots & a_{k-1} & a_k \\ 0 & 1& a_1 & \cdots & a_{k-2} & a_{k-1} \\ 0 & 0 & 1 & \cdots & a_{k-3} & a_{k-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0& 0& 0 & \cdots & 1 & a_1 \\ \end{bmatrix}.$$$$b_k = (-1)^k \det \begin{bmatrix} a_1 & a_2 & a_3 & \cdots & a_k & a_{k+1} \\ 1 & a_1 & a_2 & \cdots & a_{k-1} & a_k \\ 0 & 1& a_1 & \cdots & a_{k-2} & a_{k-1} \\ 0 & 0 & 1 & \cdots & a_{k-3} & a_{k-2} \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0& 0& 0 & \cdots & 1 & a_1 \\ \end{bmatrix}.$$
Proof sketch: The condition $\left( \sum a_k x^k \right)\left( \sum b_k x^k \right)=1$ translates into an infinite matrix condition $$\begin{bmatrix} 1 &a_1 & a_2 & a_3 & \cdots \\ 0 &1 & a_1 & a_2 & \cdots \\ 0& 0 & 1& a_1 & \cdots \\ 0& 0 & 0 & 1 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{bmatrix} \begin{bmatrix} b_0 \\ b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}. $$ Solve these equations using Cramer's rule. $\square$