Example 19 Find the maximum value $m_{n}$ of the function $f_{n}\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\frac{x_{1}}{\left(1+x_{1}+\cdots+x_{n}\right)^{2}}+$ $\frac{x_{2}}{\left(1+x_{2}+\cdots+x_{n}\right)^{2}}+\cdots+\frac{x_{n}}{\left(1+x_{n}\right)^{2}}$ (where $x_{i} \geqslant 0$). Express $m_{n}$ in terms of $m_{n-1}$, and find $\lim _{n \rightarrow \infty} m_{n}$.
Let $a_{i}=\frac{1}{1+x_{i}+\cdots+x_{n}}, 1 \leqslant i \leqslant n$, and define $a_{n+1}=1$. Then
$$1+x_{i}+x_{i+1}+\cdots+x_{n}=\frac{1}{a_{i}}$$

Also,
$$1+x_{i+1}+x_{i+2}+\cdots+x_{n}=\frac{1}{a_{i+1}}$$

Thus,
$$x_{i}=\frac{1}{a_{i}}-\frac{1}{a_{i+1}}$$

Therefore,
$$\begin{aligned}
f_{n} & =\sum_{i=1}^{n} a_{i}^{2}\left(\frac{1}{a_{i}}-\frac{1}{a_{i+1}}\right)=\sum_{i=1}^{n}\left(a_{i}-\frac{a_{i}^{2}}{a_{i+1}}\right) \\
& =\left(a_{1}+a_{2}+\cdots+a_{n}\right)-\left(\frac{a_{1}^{2}}{a_{2}}+\frac{a_{2}^{2}}{a_{3}}+\cdots+\frac{a_{n}^{2}}{1}\right)
\end{aligned}$$

To find the maximum value of $f_{n}$, construct the following inequalities:
$$\left\{\begin{array}{l}
\frac{a_{1}^{2}}{a_{2}}+\lambda_{1}^{2} a_{2} \geqslant 2 \lambda_{1} a_{1} \\
\frac{a_{2}^{2}}{a_{3}}+\lambda_{2}^{2} a_{3} \geqslant 2 \lambda_{2} a_{2} \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
\frac{a_{n}^{2}}{1}+\lambda_{n}^{2} \geqslant 2 \lambda_{n} a_{n}
\end{array}\right.$$

where $\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n}$ are parameters, $\lambda_{i} \geqslant 0$.
By adding the $n$ inequalities in (1), it is sufficient to have
$$\left\{\begin{array}{l}
2 \lambda_{1}=1 \\
2 \lambda_{2}=1+\lambda_{1}^{2} \\
\cdots \cdots \cdots \cdots \cdots \cdots \\
2 \lambda_{n}=1+\lambda_{n-1}^{2}
\end{array}\right.$$

Thus, $f_{n} \leqslant \lambda_{n}^{2}$.
Note that $\lambda_{i} \geqslant \lambda_{i-1}$, and $0 \leqslant \lambda_{i} \leqslant 1$, so $\lim _{n \rightarrow \infty} \lambda_{n}$ exists, and its value is 1.