Question 123, Let the positive sequence $\left\{a_{n}\right\}$ satisfy: $a_{1}=1, a_{n}=\frac{1}{2}\left(S_{n}+\frac{1}{S_{n}}\right)$, where $S_{n}=\sum_{i=1}^{n} a_{i}$, find the general term formula of the sequence $\left\{\mathrm{a}_{\mathrm{n}}\right\}$.
Question 123, Solution: According to the conditions, we know: $S_{1}=a_{1}=1$, and
$$
\begin{array}{l}
\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}=\mathrm{a}_{\mathrm{n}}=\frac{1}{2}\left(\mathrm{~S}_{\mathrm{n}}+\frac{1}{\mathrm{~S}_{\mathrm{n}}}\right) \\
\Rightarrow \mathrm{S}_{\mathrm{n}-1}=\frac{\mathrm{S}_{\mathrm{n}}{ }^{2}-1}{2 \mathrm{~S}_{\mathrm{n}}}
\end{array}
$$

Let $S_{n}=\cot \theta_{n}, \theta_{n} \in\left[0, \frac{\pi}{2}\right]$, then $\theta_{1}=\frac{\pi}{4}$, and $\cot \theta_{n-1}=\frac{\cot ^{2} \theta_{n}-1}{2 \cot \theta_{n}}=\cot 2 \theta_{n}$, hence $\theta_{n}=\frac{\theta_{n-1}}{2}$. Therefore, $\theta_{n}=\frac{1}{2^{n-1}} \theta_{1}=\frac{\pi}{2^{n+1}}$. Thus, $S_{n}=\cot \frac{\pi}{2^{n+1}}$, and consequently, $a_{n}=\cot \frac{\pi}{2^{n+1}}-\cot \frac{\pi}{2^{n}}$.