CF for a given sequence of approximants

The sequences $\{A_n\}_{n=-1}^\infty$ and $\{B_n\}_{n=-1}^\infty$ of complex numbers are the canonical numerators and denominators of some continued fraction $b_0+\mathbf{K}(a_n/b_n)$ if and only if

$A_{-1}=B_0=1, \quad B_{-1}=0, \quad \Delta_n=A_nB_{n-1}-B_nA_{n-1}\neq 0$

for all $n \in \mathbb{N}$ . If so, then $b_0+\mathbf{K}(a_n/b_n)$ is uniquely determined by:

$b_0=A_0, \quad b_1=B_1, \quad a_1=A_1-A_0B_1,$

$a_n={-}\frac{\Delta_n}{\Delta_{n-1}}, \quad b_n=\frac{A_nB_{n-2}-B_nA_{n-2}}{\Delta_{n-1}} \text{ for } n\ges 2.$

i.e. $\begin{tabular}{l | llll} & -1 & 0 & 1 & 2 \\ \hline A & 1 & & & \\ B & 0 & 1 & & \end{tabular}$

Example. Find a continued fraction with approximants 1/2, 3/4, 7/8, 15/16 etc.
$\begin{tabular}{l | llll} & -1 & 0 & 1 & 2 \\ \hline A & 1 & 0 & 1 & 3 \\ B & 0 & 1 & 2 & 4 \\ \hline a & & & 1 & \\ b & 0 & 0 & 2 & \end{tabular}$
$A_n=2^n-1,\quad B_n=2^n,$
$\Delta_n=(2^n-1)2^{n-1}-2^n(2^{n-1}-1)=2^{n-1},$
$\Delta_{n-1}=2^{n-2}$
$A_nB_{n-2}-B_nA_{n-2}=(2^n-1)2^{n-2}-2^n(2^{n-2}-1)=2^n-2^{n-2}=3.2^{n-2}$
for $n\ges 2$ , $\quad a_n=-2^{n-1}/2^{n-2}=-2, \quad b_n=(3.2^{n-2})/2^{n-2}=3$ .