Example 5 Find the maximum value $m_{n}$ of the function
$$
\begin{array}{l} 
f_{n}\left(x_{1}, x_{2}, \cdots, x_{n}\right) \\
= \frac{x_{1}}{\left(1+x_{1}+\cdots+x_{n}\right)^{2}}+ \\
\frac{x_{2}}{\left(1+x_{2}+\cdots+x_{n}\right)^{2}}+\cdots+\frac{x_{n}}{\left(1+x_{n}\right)^{2}}
\end{array}
$$

express $m_{n}$ in terms of $m_{n-1}$, and find $\lim _{n \rightarrow \infty} m_{n}$, where $x_{i} \geqslant 0$.
【Analysis】Noticing the linear form of the denominator, it is advisable to first make a substitution for the denominator, and then use the method of undetermined coefficients to eliminate variables to find the extremum.
Let $a_{i}=\frac{1}{1+x_{i}+\cdots+x_{n}}, a_{n+1}=1$. Then
$x_{i}=\frac{1}{a_{i}}-\frac{1}{a_{i+1}}$.
Thus, $f_{n}=\sum_{i=1}^{n} a_{i}^{2}\left(\frac{1}{a_{i}}-\frac{1}{a_{i+1}}\right)$
$$
=\sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i+1}} \text {. }
$$

Construct the following inequalities:
$$
\left\{\begin{array}{l}
\frac{a_{1}^{2}}{a_{2}}+\lambda_{1}^{2} a_{2} \geqslant 2 \lambda_{1} a_{1}, \\
\frac{a_{2}^{2}}{a_{3}}+\lambda_{2}^{2} a_{3} \geqslant 2 \lambda_{2} a_{2}, \\
\cdots \cdots \\
\frac{a_{n}^{2}}{1}+\lambda_{n}^{2} \geqslant 2 \lambda_{n} a_{n},
\end{array}\right.
$$

where, parameters $\lambda_{i} \geqslant 0$.
Adding them up, we only need to make
$$
\left\{\begin{array}{l}
2 \lambda_{1}=1, \\
2 \lambda_{2}=1+\lambda_{1}^{2}, \\
\cdots \cdots \\
2 \lambda_{n}=1+\lambda_{n-1}^{2},
\end{array}\right.
$$

which implies $f_{n} \leqslant \lambda_{n}^{2}$.
Thus, $\lambda_{n+1}^{2}=\frac{1}{4}\left(\lambda_{n}^{2}+1\right)^{2}$.
Therefore, $\lim _{n \rightarrow \infty} \lambda_{n}^{2}=1$.