Three. (20 points) Given a sequence $\left\{a_{n}\right\}$ with all terms no less than 1, satisfying: $a_{1}=1, a_{2}=1+\frac{\sqrt{2}}{2},\left(\frac{a_{n}}{a_{n+1}-1}\right)^{2}+$ $\left(\frac{a_{n}-1}{a_{n-1}}\right)^{2}=2$. Try to find:
(1) The general term formula of the sequence $\left\{a_{n}\right\}$;
(2) The value of $\lim _{n \rightarrow+\infty} \frac{a_{n}}{n}$.
(1) Let $b_{n}=\left(\frac{a_{n+1}-1}{a_{n}}\right)^{2}$.
Then $\frac{1}{b_{n}}+b_{n-1}=2$, and $b_{1}=\frac{1}{2}$.
From this, $b_{2}=\frac{2}{3}, b_{3}=\frac{3}{4}, \cdots$.
Observing, we find $b_{n}=\frac{n}{n+1}$.
We will prove this by mathematical induction.
For $n=1,2,3$, the conclusion is obviously true.
Assume that for $n=k$, the conclusion holds, and we will prove that for $n=k+1$, the conclusion also holds.
From $b_{k}=\frac{k}{k+1}$, we know $b_{k+1}=\frac{1}{2-b_{k}}=\frac{k+1}{k+2}$.
Therefore, $b_{n}=\frac{n}{n+1}$ holds for all positive integers $n$.

From this, we get $\left(a_{n+1}-1\right)^{2}=\frac{n}{n+1} a_{n}^{2}$.
Given $a_{n} \geqslant 1$, we have $a_{n+1}=\sqrt{\frac{n}{n+1}} a_{n}+1$.
Let $c_{n}=\sqrt{n} a_{n}$, then $c_{n+1}=c_{n}+\sqrt{n+1}$, and $c_{1}=1$.
Thus, $c_{n}=c_{n-1}+\sqrt{n}=c_{n-2}+\sqrt{n}+\sqrt{n-1}$
$$
=\cdots=\sum_{k=1}^{n} \sqrt{k} \text {. }
$$

Therefore, $a_{n}=\frac{1}{\sqrt{n}} \sum_{k=1}^{n} \sqrt{k}$.
(2) Notice that
$$
\begin{array}{l}
\sum_{k=1}^{n} \sqrt{k}>\sum_{k=1}^{n} \frac{2 k}{\sqrt{k}+\sqrt{k+1}} \\
=2 \sum_{k=1}^{n} k(\sqrt{k+1}-\sqrt{k}) \\
=2\left(-\sum_{k=1}^{n} \sqrt{k}+n \sqrt{n+1}\right) \text { (Abel's formula), } \\
\text { hence } \sum_{k=1}^{n} \sqrt{k}>\frac{2}{3} n \sqrt{n+1}>\frac{2}{3} n \sqrt{n}, \text { i.e., } \frac{a_{n}}{n}>\frac{2}{3} .
\end{array}
$$

Thus, $\sum_{k=1}^{n} \sqrt{k}>\frac{2}{3} n \sqrt{n+1}>\frac{2}{3} n \sqrt{n}$, i.e., $\frac{a_{n}}{n}>\frac{2}{3}$.
Therefore, $\lim _{n \rightarrow+\infty} \frac{a_{n}}{n} \geqslant \frac{2}{3}$.
Similarly, since
$$
\begin{array}{l}
\sum_{k=1}^{n} \sqrt{k}<\sum_{k=1}^{n} \frac{2 k}{\sqrt{k-1}+\sqrt{k}} \\
=2 \sum_{k=1}^{n} k(\sqrt{k}-\sqrt{k-1}) \\
=2\left[(n+1) \sqrt{n}-\sum_{k=1}^{n} \sqrt{k}\right], \\
\text { hence } \sum_{k=1}^{n} \sqrt{k}<\frac{2(n+1)}{3} \sqrt{n}, \text { i.e., } \\
\frac{a_{n}}{n}<\frac{2}{3}\left(1+\frac{1}{n}\right) .
\end{array}
$$

Thus, $\sum_{k=1}^{n} \sqrt{k}<\frac{2(n+1)}{3} \sqrt{n}$, i.e.,

Therefore, $\lim _{n \rightarrow+\infty} \frac{a_{n}}{n} \leqslant \frac{2}{3}\left(1+\lim _{n \rightarrow+\infty} \frac{1}{n}\right)=\frac{2}{3}$.
In summary, $\lim _{n \rightarrow+\infty} \frac{a_{n}}{n}=\frac{2}{3}$.