8. Let $n$ be a positive integer. For positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have $a_{1}+a_{2}+\cdots+a_{n}=1$. Let $A$ be the minimum of the following $n$ numbers $\frac{a_{1}}{1+a_{1}}, \frac{a_{2}}{1+a_{1}+a_{2}}, \cdots, \frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n}}$. Find the maximum value of $A$ when $a_{1}$, $a_{2}, \cdots, a_{n}$ are variables. (2004 Japan Mathematical Olympiad Problem)
8. From $A=\min _{1 \leqslant i \leqslant n}\left\{\frac{a_{i}}{1+\sum_{k=1}^{i} a_{k}}\right\}$, we know that $1-A=\max _{1 \leqslant i \leqslant n}\left\{\frac{1+\sum_{k=1}^{i-1}}{1+\sum_{k=1}^{i} a_{k}}\right\} \geqslant \sqrt[n]{\prod_{i=1}^{n} \frac{1+\sum_{k=1}^{i-1}}{1+\sum_{k=1}^{i} a_{k}}}=$ $\sqrt[n]{\frac{1}{1+\sum_{k=1}^{i} a_{k}}}=\sqrt[n]{\frac{1}{2}}$. Therefore, $A \leqslant 1-\frac{1}{\sqrt[n]{2}}$. The equality holds if and only if $\frac{a_{i}}{1+\sum_{k=1}^{i} a_{k}}=1-\frac{1}{\sqrt[n]{2}}$, i.e., $a_{i}=$ $\sqrt[n]{2^{i-1}}(\sqrt[n]{2}-1)(i=1,2, \cdots, n)$.