Example 4 Let $p, q$ be the roots of the quadratic equation
$$
x^{2}+2 a x-1=0(a>0)
$$

where $p>0$. Let
$$
y_{1}=p-q, y_{n+1}=y_{n}^{2}-2(n=1,2, \cdots) .
$$
Prove: $\lim _{n \rightarrow \infty}\left(\frac{1}{y_{1}}+\frac{1}{y_{1} y_{2}}+\cdots+\frac{1}{y_{1} y_{2} \cdots y_{n}}\right)=p$.
(2010, Joint Autonomous Admissions Examination of Tsinghua University and Other Schools)
By Vieta's formulas, we know
$$
\begin{array}{l}
p q=-1, y_{1}=p-q=p+\frac{1}{p}, \\
y_{2}=y_{1}^{2}-2=p^{2}+\frac{1}{p^{2}}, \cdots
\end{array}
$$

By mathematical induction, we can prove
$$
\begin{array}{l} 
y_{n}=p^{2 n-1}+\frac{1}{p^{2 n-1}} . \\
\text { Hence } \sum_{k=1}^{n}\left(\frac{1}{\prod_{i=1}^{k} y_{i}}\right)=\sum_{k=1}^{n} \prod_{i=1}^{k} \frac{p^{2 i-1}}{1+p^{2^{i}}} \\
=\frac{1}{p} \sum_{k=1}^{n}\left(\prod_{i=1}^{k-1} \frac{1}{1+p^{2 i}}-\prod_{i=1}^{k} \frac{1}{1+p^{2 i}}\right) \\
=\frac{1}{p}\left(1-\frac{1-p^{2}}{1-p^{2 n+1}}\right) .
\end{array}
$$

By $p>0$, we get
$$
\begin{array}{l}
p=\sqrt{a^{2}+1}-a=\frac{1}{\sqrt{a^{2}+1}+a}<1 . \\
\text { Hence } \lim _{n \rightarrow \infty}\left(\frac{1}{y_{1}}+\frac{1}{y_{1} y_{2}}+\cdots+\frac{1}{y_{1} y_{2} \cdots y_{n}}\right)=p .
\end{array}
$$