The Value of a Finite Continued Fraction
Let
\[[ a_1,a_2,...,a_{n-1},a_n ]\]
be a finite continued fraction. Define the numerators \[p_1, \; p_2,..., \; p_{n-1}, \; p_n\]
and the denominators \[q_1, \; q_2,..., \; q_{n-1}, \; q_n\]
of the convergents by\[p_1=a_1, \; p_2=a_1a_2+1, \; p_k=a_kp_{k-1}+p_{k-2}, \; 1 \le 3 k \le n\]
\[q_1=1, \; q_2=a_2, \; q_k=a_kq_{k-1}+q_{k-2}, \; 1 \le 3 k \le n\]
Then the value of the finite continued fraction is
\[\frac{p_n}{q_n}\]
.Proof
Proof is by induction. Let
\[P(k)\]
be the statement that the value of \[[ a_1,a_2,...,a_{k-1},a_k ]\]
is \[\frac{p_k}{q_k}\]
, with \[p_k, \; q_k\]
as defined above.Then
\[P(1)\]
is true since \[[ a_1 ] = \frac{a_1}{1}= \frac{p_1}{q_1}\]
.\[[ a_1, a_2 ] =a_1 + \frac{1}{a_2} = \frac{a_1a_2+1}{a_2}= \frac{p_2}{q_2}\]
so \[P(2)\]
is true. Suppose then that
\[P(k)\]
is true for all integers up to and including \[k\]
. Consider \[[ a_1,a_2,...,a_{k-1},a_k, a_{k+1} ]\]
. The numerators of the convergents are \[p_1, \; p_2,..., \; p_{k-1}, \; p_k, \; p_{k+1} \]
and the denominators are \[q_1, \; q_2,..., \; q_{k-1}, \; q_k, \; q_{k+1} \]
.By The Second Continued Fraction Identity,
\[[ a_1,a_2,...,a_k,a_{k+1} ]= [ a_1,a_2,...,a_k+\frac{1}{a_{k+1}} ]\]
.The first
\[k\]
numerators of the convergents of this continued fraction are \[p_1, \; p_2,..., p_{k-1}, \; p'_k=a_k+\frac{1}{a_{k+1}}p_{k-1}+p_{k-2}\]
.The first
\[k\]
denominators are \[q_1, \; q_2,..., q_{k-1}, \; q'_k=a_k+\frac{1}{a_{k+1}}q_{k-1}+q_{k-2}\]
.The kth convergent is
\[\frac{p_k}{q_k}\]
.\[ \begin{equation} \begin{aligned} \left[ a_1,a_2,...,a_k,a_{k+1} \right] &= \frac{p'_k}{q'_k} \\ &= \frac{(a_k+\frac{1}{a_{k+1}})p_{k-1}+p_{k-2}}{(a_k+\frac{1}{a_{k+1}})q_{k-1}+q_{k-2}} \\ &= \frac{a_{k+1}(a_kp_{k-1}+p_{k-2})+p_k}{a_{k+1}(a_kq_{k-1}+q_{k-2})+q_k} \\ &= \frac{a_{k+1}p_k+p_{k-1}}{a_{k+1}q_k+q_{k-1}}\\ &= \frac{p_{k+1}}{q_{k+1}} \end{aligned} \end{equation}\]
