Let \(A(z)=\sum_{k\ge0}c_kz^k\) be a power series with positive radius of convergence and \(c_0\neq0\). Then \(1/A\) is holomorphic near \(0\), and its Taylor series \(1/A(z)=\sum_{k\ge0}b_kz^k\) has coefficients determined recursively by \[ b_0=\inverseof{c_0},\qquad b_n=-\inverseof{c_0}\sum_{k=1}^{n}c_kb_{n-k}\quad(n\ge1). \] In particular, \(b_1=-\inverseof{c_0} c_1b_0\) and \(b_2=-\inverseof{c_0}(c_2b_0+c_1b_1)\).
Since \(A(0)=c_0\neq0\) and \(A\) is continuous, \(A\) has no zeros near \(0\), so \(1/A\) is holomorphic there and has a Taylor series \(\sum_k b_kz^k\). The Cauchy product of the two series is \(A(z)\cdot\frac{1}{A(z)}=1\), so comparing coefficients gives \(c_0b_0=1\) and \(\sum_{k=0}^{n}c_kb_{n-k}=0\) for \(n\ge1\). Solving the last equation for \(b_n\) gives the recursion.