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.

So the sought continued fraction is \cfrac{1}{2-\cfrac{2}{3-\cfrac{2}{3-\cfrac{2}{3-\dots}}}}

references

Lorentzen, Waadeland – Continued fractions with applications

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