2. (2nd Quiz, 2nd Question, provided by Huang Yumin) Given a sequence of positive numbers $a_{1}, a_{2}, a_{3}, \cdots$, satisfying $a_{n+1}$ $=\frac{1}{a_{1}+a_{2}+\cdots+a_{n}}(n \in N)$. Find $\lim _{n \rightarrow \infty} \sqrt{n} a_{n}$.
Solve: First, prove a limit equality:
$$
\lim _{n \rightarrow \infty} \frac{1}{n}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)=0 \text {. }
$$

Let \( T_{n}=\sum_{k=1}^{n} \frac{1}{k}, b_{n}=\frac{1}{n} T_{n} \). Then
$$
\begin{aligned}
T= & 1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right. \\
& \left.+\frac{1}{8}\right)+\cdots+\left(\frac{1}{2^{n-1}+1}+\cdots+\frac{1}{2^{n}}\right)
\end{aligned}
$$

From (2), we immediately have
$$
\begin{aligned}
T_{2} & >1+\frac{1}{2}+2 \cdot \frac{1}{4}+1 \cdot \frac{1}{8}+\cdots+2^{-1} \cdot \frac{1}{2^{2}} \\
& =1+\frac{n}{2}
\end{aligned}
$$

and \( T_{2^{*}2(n-1)} \).

From (15) and (14), we have
$$
\begin{aligned}
S_{n}^{z} & <2(n-1)+\left(a_{1}{ }^{2}+\frac{1}{a_{1}{ }^{2}}\right)+\frac{1}{2}\left(1+\frac{1}{2}\right. \\
& \left.+\cdots+\frac{1}{n-2}\right) .
\end{aligned}
$$

From (15) and (16), we have
$$
\begin{aligned}
2\left(1-\frac{1}{n}\right) & <\frac{1}{n} S_{n}^{2} \\
& <2\left(1-\frac{1}{n}\right)+\frac{1}{n}\left(a_{1}{ }^{2}+\frac{1}{a_{2}^{2}}\right. \\
& +\frac{1}{2 n}\left(1+\frac{1}{2}+\cdots+\frac{1}{n-2}\right)
\end{aligned}
$$

From (1), we immediately have
$$
\lim _{n \rightarrow \infty} \frac{1}{2 n}\left(1+\frac{1}{2}+\cdots+\frac{1}{n-2}\right)=0 .
$$

Clearly, \(\lim _{n \rightarrow \infty} 2\left(1-\frac{1}{n}\right)=2, \lim _{n \rightarrow \infty} \frac{1}{n}\left(a_{1}{ }^{2}+\frac{1}{a_{1}{ }^{2}}\right)=0\). Therefore, \(\lim _{n \rightarrow \infty} \frac{1}{n} S_{n}^{2}=2\).

Hence, \(\lim _{n \rightarrow \infty} \sqrt{n} a_{n}=\lim _{n \rightarrow \infty} \sqrt{n} \frac{1}{S_{n-1}}\)
$$
=\lim _{n \rightarrow \infty} \frac{\sqrt{n}}{\sqrt{n-1}} \lim _{n \rightarrow \infty} \frac{\sqrt{n-1}}{S_{n-1}}=\frac{\sqrt{2}}{2} .
$$