(related to Lemma: Continuants and Convergents)
By induction, the above definition of continuants, as well as convergents:
$[q_0;q_1]=q_0+\frac{1}{q_1}=q_0+\frac{Q_0}{Q_1(q_1)}$
$$\begin{array}{rcll} [q_0;q_1,\ldots,q_{k+1}]&=&q_0+\frac{1}{[q_1;q_2,\ldots,q_k]}\\ &=&q_0+\cfrac{1}{q_1+\cfrac{Q_{k-1}(q_2,\ldots,q_{k+1})}{Q_k(q_2,\ldots,q_{k+1})}}&\text{by base case}\\ &=&q_0+\cfrac{Q_k(q_2,\ldots,q_{k+1})}{q_1Q_k(q_2,\ldots,q_{k+1})+Q_{k-1}(q_3,\ldots,q_{k+1})}. \end{array}$$
