Example 2 Let $A B C D$ be a trapezoid, $A D / / B C, A D=a, B C=b$, the diagonals $A C$ and $B D$ intersect at $P_{1}$, draw $P_{1} Q_{1} / / A D$ intersecting $C D$ at $Q_{1}$, connect $A Q_{1}$, intersecting $B D$ at $P_{2}$, draw $P_{2} Q_{2} / / A D$ intersecting $C D$ at $Q_{2}$, connect $A Q_{2}$ intersecting $B D$ at $P_{3}, \cdots$, if this continues, let the length of $P_{n} Q_{n}$ be $x_{n}$.
(1) Express $x_{n}$ in terms of $a, b, n$; (2) Prove that $x_{n} \leqslant \frac{a+n b}{4 n}$.
Solution: (1) As shown in Figure 4-2 (a), since $A D=a$, $B C=b$, according to the problem, we have
$$
P_{1} Q_{1}=x_{1}=\frac{a b}{a+b},
$$

Taking out the trapezoid $A P_{k} Q_{k} D$, as shown in Figure (b), we get
$$
\begin{array}{l}
\frac{x_{k+1}}{x_{k}}=\frac{D Q_{k+1}}{D Q_{k}} . \\
\text { Also } \quad \frac{x_{k+1}}{A D}+\frac{D Q_{k+1}}{D Q_{k}}=1,
\end{array}
$$

Therefore, $\frac{x_{k+1}}{x_{k}}+\frac{x_{k+1}}{A D}=1$,
which means $\frac{x_{k+1}}{x_{k}}+\frac{x_{k+1}}{a}=1, \frac{1}{x_{k+1}}-\frac{1}{x_{k}}=\frac{1}{a}$,
so the sequence $\left\{\frac{1}{x_{k}}\right\}$ is an arithmetic sequence with a common difference of $\frac{1}{a}$, thus $\frac{1}{x_{n}}=\frac{a+b}{a b}+\frac{n-1}{a}=\frac{a+n b}{a b}$,
hence $x_{n}=\frac{a b}{a+n b}$.
(2) $x_{n}=\frac{a b}{a+n b}=\frac{a \cdot n b}{n(a+n b)} \leqslant \frac{\left(\frac{a+n b}{2}\right)^{2}}{n(a+n b)}=\frac{a+n b}{4 n}$,

Therefore, $x_{n} \leqslant \frac{a+n b}{4 n}$.