Example 7 Convert $\sqrt{11}$ into a continued fraction, and find its first four convergents.
Explanation: Since the largest integer not exceeding $\sqrt{11}$ is 3, we set $\sqrt{11}=3+\frac{1}{a_{1}}\left(0<\frac{1}{a_{1}}<1\right)$. Then
$$
\begin{array}{l}
a_{1}=\frac{1}{\sqrt{11}-3}=\frac{\sqrt{11}+3}{2} \\
=3+\frac{1}{a_{2}}\left(0<\frac{1}{a_{2}}<1\right), \\
a_{2}=\frac{2}{\sqrt{11}-3}=\sqrt{11}+3 \\
=6+\frac{1}{a_{3}}\left(0<\frac{1}{a_{3}}<1\right), \\
a_{3}=\frac{1}{\sqrt{11}-3}=a_{1} .
\end{array}
$$

Therefore, $\sqrt{11}=3+\frac{1}{3}+\frac{1}{6}+\frac{1}{3}+\cdots$
$$
=[3, \overline{3,6}] \text {. }
$$

Thus, we have Table 6.
Table 6
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline$a_{n}$ & & & 3 & 3 & 6 & 3 & 6 & $\cdots$ \\
\hline$p_{n}$ & 0 & 1 & 3 & 10 & 63 & 199 & 1257 & $\cdots$ \\
\hline$q_{n}$ & 1 & 0 & 1 & 3 & 19 & 60 & 379 & $\cdots$ \\
\hline
\end{tabular}

Its first four convergents are $\frac{3}{1}$, $\frac{10}{3}$, $\frac{63}{19}$, and $\frac{199}{60}$, and the error between $\frac{199}{60}$ and $\frac{63}{19}$ is less than $\frac{1}{60^{2}}$.