3 Let $x_{1}, x_{2}, \cdots, x_{n}$ be non-negative real numbers, and denote $H=\frac{x_{1}}{\left(1+x_{1}+x_{2}+\cdots+x_{n}\right)^{2}}+$ $\frac{x_{2}}{\left(1+x_{2}+x_{3}+\cdots+x_{n}\right)^{2}}+\cdots+\frac{x_{n}}{\left(1+x_{n}\right)^{2}}$ with the maximum value as $a_{n+1}$. When do $x_{1}$, $\dot{x}_{2}, \cdots, x_{n}$ achieve the maximum value of $H$? And find the relationship between $a_{n}$ and $a_{n-1}$, as well as $\lim _{n \rightarrow \infty} a_{n}$.
3. First prove the lemma: Let $g(x)=\frac{a}{x+b}+\frac{x}{(x+b)^{2}}$, where $a \geqslant 0, b \geqslant 1$. Using the discriminant method, it can be shown that when $x=\frac{b(1-a)}{1+a}$, the maximum value of $g(x)$ is $\frac{(1+a)^{2}}{4 b}$. Original problem solution: Fix $x_{2}, x_{3}, \cdots, x_{n}$, then $H$ is a function of $x_{1}$, $g\left(x_{1}\right)+C_{1}$, where $a_{1}=0$, $b_{1}=1+x_{2}+\cdots+x_{n}$. By the lemma, when $x_{1}=\frac{\left(1+x_{2}+x_{3}+\cdots+x_{n}\right)\left(1-a_{1}\right)}{1+a_{1}}$, the maximum value of $g\left(x_{1}\right)+C_{1}$ is: $\frac{\left(1+a_{1}\right)^{2}}{4} \cdot \frac{1}{1+x_{2}+x_{3}+\cdots+x_{n}}+$ $\frac{x_{2}}{\left(1+x_{2}+x_{3}+\cdots+x_{n}\right)^{2}}+\cdots+\frac{x_{n}}{\left(1+x_{n}\right)^{2}}=H_{2}$. Fix $x_{3}, x_{4}, \cdots, x_{n}$, then $H_{2}$ is a function of $x_{2}$, $g\left(x_{2}\right)+C_{2}$, where $a_{2}=\frac{\left(1+a_{1}\right)^{2}}{4}, b_{2}=1+x_{3}+\cdots+x_{n}$. By the lemma, when $x_{2}=\frac{\left(1+x_{3}+x_{4}+\cdots+x_{n}\right)\left(1-a_{2}\right)}{1+a_{2}}$, the maximum value of $g\left(x_{2}\right)+C_{2}$ is $\frac{\left(1+a_{2}\right)^{2}}{4} \cdot \frac{1}{1+x_{3}+x_{4}+\cdots+x_{n}}+\frac{x_{3}}{\left(1+x_{3}+x_{4}+\cdots+x_{n}\right)^{2}}+\cdots+$ $\frac{x_{n}}{\left(1+x_{n}\right)^{2}}=H_{3}$. Continuing this process, we get $\frac{\left(1+a_{n-1}\right)^{2}}{4} \cdot \frac{1}{1+x_{n}}+\frac{x_{n}}{\left(1+x_{n}\right)^{2}}=H_{n}$. Using the lemma (when $b=1$), when $x_{n}=\frac{1-a_{n}}{1+a_{n}}$, the maximum value of $H_{n}$ is $\frac{\left(1+a_{n}\right)^{2}}{4}$. Here, $a_{n}=$ $\frac{\left(1+a_{n-1}\right)^{2}}{4}$. Therefore, if $a_{n+1}$ is the maximum value of $H$, then $a_{n}$ satisfies: $a_{1}=0, a_{k}=$ $\frac{\left(1+a_{k-1}\right)^{2}}{4}$. The maximum value is achieved when $x_{n}=\frac{1-a_{n}}{1+a_{n}}, x_{n-1}=\frac{\left(1+x_{n}\right)\left(1-a_{n-1}\right)}{1+a_{n-1}}, \cdots, x_{1}=$ $\frac{\left(1+x_{2}+x_{3}+\cdots+x_{n}\right)\left(1-a_{1}\right)}{1+a_{1}}$. It is easy to see that $a_{n} \geqslant a_{n-1}$, and when $0 \leqslant a_{n-1} \leqslant 1$, $0 \leqslant a_{n} \leqslant 1$. Therefore, $x_{1}, x_{2}, \cdots, x_{n}$ are all non-negative. Note that $a_{n}$ is a monotonic and bounded sequence, so it must have a limit. Let the limit be $a$, then $a=\frac{(1+a)^{2}}{4}$, i.e., $a=1$.