Example 11 In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, and $a_{n}=$ $1+\frac{1}{a_{n-1}}, n=2,3, \cdots$. Find $a_{n}$.
Given: $f(x)=1+\frac{1}{x}$ has two distinct fixed points
$$
x_{1}=\frac{1}{2}(1+\sqrt{5}), \quad x_{2}=\frac{1}{2}(1-\sqrt{5})
$$
(From the definition of a fixed point of a function, we have $1+\frac{1}{x}=x$, solving which gives these results). By the conclusion of Theorem 2(1), we have
$$
\begin{array}{l}
\frac{a_{n}-\frac{1+\sqrt{5}}{2}}{a_{n}-\frac{1-\sqrt{5}}{2}}=\left(\frac{A-C x_{1}}{A-C x_{2}}\right) \cdot \frac{a_{1}-\frac{1+\sqrt{5}}{2}}{a_{1}-\frac{1-\sqrt{5}}{2}} \\
=\left(\frac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n} \text { (where, } A=1, C=1 \text { ). }
\end{array}
$$

Solving this equation for $a_{n}$, we get
$$
a_{n}=\frac{1}{2}\left[\frac{(1+\sqrt{5})^{n+1}-(1-\sqrt{5})^{n+1}}{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}\right], n \in \mathbf{N} \text {. }
$$