Example 2 Find the finite simple continued fraction of $7700 / 2145$ and its successive convergents.
Solve
$$\begin{aligned}
7700 & / 2145=\langle 3+1265 / 2145\rangle=\langle 3,2145 / 1265\rangle \\
& =\langle 3,1+880 / 1265\rangle=\langle 3,1,1265 / 880\rangle \\
= & \langle 3,1,1+385 / 880\rangle=\langle 3,1,1,880 / 385\rangle \\
= & \langle 3,1,1,2+110 / 385\rangle=\langle 3,1,1,2,385 / 110\rangle \\
= & \langle 3,1,1,2,3+55 / 110\rangle=\langle 3,1,1,2,3,2\rangle \\
= & \langle 3,1,1,2,3,1,1\rangle
\end{aligned}$$

According to formula (2) in §2, list to find $h_{n}, k_{n}$ (see Table 1) and the convergents $h_{n} / k_{n}$. From Table 1, the convergents are sequentially
$$3 / 1,4 / 1,7 / 2,18 / 5,61 / 17,140 / 39=7700 / 2145$$

Table 1
\begin{tabular}{|c|rrrrrr|}
\hline$n$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline$a_{n}$ & 3 & 1 & 1 & 2 & 3 & 2 \\
$h_{n}$ & 3 & 4 & 7 & 18 & 61 & 140 \\
$k_{n}$ & 1 & 1 & 2 & 5 & 17 & 39 \\
\hline
\end{tabular}

This fraction simplifies to $140 / 39$. Although the original fraction is not in its simplest form, the convergents are always in their simplest form. Using formulas (7) and (8), the error between the original fraction and the convergents can be calculated, such as
$$\begin{aligned}
140 / 39-61 / 17 & =h_{5} / k_{5}-h_{4} / k_{4}=1 /\left(k_{4} k_{5}\right) \\
& =1 /(17 \cdot 39)=1 / 663 \\
140 / 39-18 / 5 & =h_{5} / k_{5}-h_{3} / k_{3}=-a_{5} /\left(k_{3} k_{5}\right) \\
& =-2 /(5 \cdot 39)=-2 / 195
\end{aligned}$$

The other errors are left for the reader to calculate.