# 連分數

convergents of the continued fraction expansion of $\frac{13}{17}$

\begin{aligned} c_0 = 0 = \frac{0}{1} \\ c_1 = 0 + \frac{1}{1} = \frac{1}{1} \\ c_2 = 0 + \frac{1}{1 + \frac{1}{3}} = \frac{3}{4} \\ c_3 = 0 + \frac{1}{1 + \frac{1}{3 + \frac{1}{4}}} = \frac{13}{17} \end{aligned}