Example 2 Find the convergent fraction of $3+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=[3,2,3, 4,5]$.
Explanation: First, list $p_{-1}=0, p_{0}=1, q_{-1}=1, q_{0}=$ $0, a_{1}=3, a_{2}=2, a_{3}=3, a_{4}=4, a_{5}=5$, in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline$n$ & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline$a_{n}$ & & & 3 & 2 & 3 & 4 & 5 \\
\hline$p_{n}$ & 0 & 1 & & & & & \\
\hline$q_{n}$ & 1 & 0 & & & & & \\
\hline
\end{tabular}

Then, use the recurrence relations in Theorem 1 to sequentially calculate the values of $p_{n}$ and $q_{n}$ and fill them into Table 2.
Table 2
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline$n$ & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline$a_{n}$ & & & 3 & 2 & 3 & 4 & 5 \\
\hline$p_{n}$ & 0 & 1 & 3 & 7 & 24 & 103 & 539 \\
\hline$q_{n}$ & 1 & 0 & 1 & 2 & 7 & 30 & 157 \\
\hline
\end{tabular}

Thus, the continued fraction $[3,2,3,4,5]$ has the following convergents: $\frac{3}{1}, \frac{7}{2}, \frac{24}{7}, \frac{103}{30}, \frac{539}{157}$.

Theorem 2 states that the odd-numbered convergents are less than all subsequent convergents, and the even-numbered convergents are greater than all subsequent convergents.

Since $\frac{p_{1}}{q_{1}}=a_{1}+\frac{1}{a_{2}+\cdots}, \frac{p_{2}}{q_{2}}$ is greater than all subsequent convergents;
$$
\frac{p_{3}}{q_{3}}=a_{1}+\frac{1}{a_{2}+\frac{1}{a_{3}}}<a_{1}+\frac{1}{a_{2}+\frac{1}{a_{3}+\cdots}},
$$
$\frac{p_{3}}{q_{3}}$ is less than all subsequent convergents;
$$
\begin{array}{l}
\cdots \cdots \\
\text { Hence } \frac{p_{1}}{q_{1}}<\frac{p_{3}}{q_{3}}<\frac{p_{5}}{q_{5}}<\cdots<\frac{p_{2 n-1}}{q_{2 n-1}}<\cdots \\
<\frac{p_{2 n}}{q_{2 n}}<\cdots<\frac{p_{6}}{q_{6}}<\frac{p_{4}}{q_{4}}<\frac{p_{2}}{q_{2}}.
\end{array}
$$

For example, for the convergents of $[3,2,3,4,5]$: $\frac{3}{1}, \frac{7}{2}, \frac{24}{7}, \frac{103}{30}, \frac{539}{157}$, we have $\frac{3}{1}<\frac{24}{7}<\frac{539}{157}<\frac{103}{30}<\frac{7}{2}$.

Theorem 3 states that the numerators and denominators of two consecutive convergents satisfy
$$
p_{n} q_{n-1}-p_{n-1} q_{n}=(-1)^{n}.
$$

For example, for the convergents of $[3,2,3,4,5]$: $\frac{3}{1}, \frac{7}{2}, \frac{24}{7}, \frac{103}{30}, \frac{539}{157}$,
$$
\begin{array}{l}
7 \times 1-3 \times 2=1, \\
24 \times 2-7 \times 7=-1, \\
103 \times 7-24 \times 30=1, \\
539 \times 30-103 \times 157=-1.
\end{array}
$$

Corollary 1 states that each convergent is a reduced fraction.
For example, $\frac{3}{1}, \frac{7}{2}, \frac{24}{7}, \frac{103}{30}, \frac{539}{157}$ are all reduced fractions.
Corollary 2 states that the difference between two consecutive convergents satisfies
$$
\frac{p_{n}}{q_{n}}-\frac{p_{n-1}}{q_{n-1}}=\frac{(-1)^{n}}{q_{n} q_{n-1}}.
$$

For example, $\frac{7}{2}-\frac{3}{1}=\frac{1}{1 \times 2}$,
$$
\begin{array}{l}
\frac{24}{7}-\frac{7}{2}=\frac{-1}{2 \times 7}, \\
\frac{103}{30}-\frac{24}{7}=\frac{1}{7 \times 30}, \\
\frac{539}{157}-\frac{103}{30}=\frac{-1}{30 \times 157}.
\end{array}
$$