# Thread: Convergent in Continued fraction

1. ## Convergent in Continued fraction

If $\frac{p_k}{q_k}$ is the $k$th convergent of $[a_0;a_1, ... ,a_n]$ and $a_0 >0$, show

$\frac{q_k}{q_{k-1}} = [a_k:a_{k-1}, ... , a_2, a_1]$.

I think i have to use recurrence relation of the continued fraction: $q_k=a_kq_{k-1}+q_{k-2}$ but I don't know how to.

Can anyone help me on the question. Thank you.

2. Originally Posted by kleenex
If $\frac{p_k}{q_k}$ is the $k$th convergent of $[a_0;a_1, ... ,a_n]$ and $a_0 >0$, show

$\frac{q_k}{q_{k-1}} = [a_k:a_{k-1}, ... , a_2, a_1]$.

I think i have to use recurrence relation of the continued fraction: $q_k=a_kq_{k-1}+q_{k-2}$ but I don't know how to.

Can anyone help me on the question. Thank you.
You do this by induction. Note that, $\frac{q_k}{q_{k-1}} = \frac{a_kq_{k-1}+q_{k-2}}{q_{k-1}} = a_k + \frac{q_{k-2}}{q_{k-1}}$.
But $\frac{q_{k-1}}{q_{k-2}} = [a_{k-1};a_{k-2},...,a_1]$ thus, $\frac{q_{k-2}}{q_{k-2}} = [0;a_{k-1},a_{k-2},...,a_1]$. And so $a_k + \frac{q_{k-2}}{q_{k-1}} = a_k + [0;a_{k-1},...,a_1] = [a_k;a_{k-1},...,a_1]$.