By writing a continued fraction $[q_0;q_1,q_2,\ldots]$ as a usual rational number, we get the results
$$\begin{array}{rcl} [q_0]&=&q_0\\ [q_0;q_1]&=&q_0+\frac{1}{q_1}\\ [q_0;q_1,q_2]&=&q_0+\cfrac{1}{q_1+\cfrac{1}{q_2}}=q_0+\frac{q_2}{q_1q_2+1}\\ [q_0;q_1,q_2,q_3]&=&q_0+\cfrac{1}{q_1+\cfrac{1}{q_2+\cfrac{1}{q_3}}}=q_0+\frac{1}{q_1+\frac{q_3}{q_2q_3+1}}=\frac{q_2q_3+1}{q_1q_2q_3+q_1+q_3}\\ \end{array}$$ This motivates the following definition and lemma:
Let $[q_0;q_1,q_2,\ldots,]$ be a given continued fraction. Then we can write the $k$-th convergent as $$[q_0;q_1,q_2,\ldots,q_k]=q_0+\frac{Q_{k-1}(q_2,\ldots,q_k)}{Q_{k}(q_1,q_2,\ldots,q_k)}$$ with the so-called $k$-th continuant function $Q:\mathbb Z^k\to\mathbb Z$ defined recursively by $$Q_k(x_1,\ldots,x_k):=\begin{cases}0&\text{if }k=-1,\\1&\text{if }k=0,\\x_1&\text{if }k=1,\\q_1Q_{k-1}(x_2,\ldots,x_k)+Q_{k-2}(x_3,\ldots,x_k)&\text{else.}\end{cases}$$
Proofs: 1
