Given a sequence $\left\{a_{n}\right\}$ with all terms no less than 1, it satisfies:
$$
a_{1}=1, \quad a_{2}=1+\frac{\sqrt{2}}{2}, \quad\left(\frac{a_{n}}{a_{n+1}-1}\right)^{2}+\left(\frac{a_{n}-1}{a_{n-1}}\right)^{2}=2 .
$$

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 that $b_{n}=\frac{n}{n+1}$. We will prove this by mathematical induction.
When $n=1,2,3$, the conclusion is obviously true. Assume that the conclusion holds for $n=k$, we need to prove that it also holds for $n=k+1$. Given $b_{k}=\frac{k}{k+1}$, we have $b_{k+1}=\frac{1}{2-b_{k}}=\frac{k+1}{k+2}$. Therefore, $b_{n}=\frac{n}{n+1}$ holds for all positive integers $n$.
From this, we get $\left(a_{n+1}-1\right)^{2}=\frac{n}{n+1} a_{n}^{2}$. Since $a_{n} \geqslant 1$, 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 $c_{1}=1$. Hence,
$$
c_{n}=c_{n-1}+\sqrt{n}=c_{n-2}+\sqrt{n}+\sqrt{n-1}=\cdots=\sum_{k=1}^{n} \sqrt{k} .
$$

Thus, $a_{n}=\frac{1}{\sqrt{n}} \sum_{k=1}^{n} \sqrt{k}$.
(2) By Abel's formula, we get
$$
\sum_{k=1}^{n} \sqrt{k}>\sum_{k=1}^{n} \frac{2 k}{\sqrt{k}+\sqrt{k+1}}=2 \sum_{k=1}^{n} k(\sqrt{k+1}-\sqrt{k})=2\left(-\sum_{k=1}^{n} \sqrt{k}+n \sqrt{n+1}\right) \text {, }
$$

Thus,
$$
\sum_{k=1}^{n} \sqrt{k}>\frac{2}{3} n \sqrt{n+1}>\frac{2}{3} n \sqrt{n}
$$

which implies $\frac{a_{n}}{n}>\frac{2}{3}$. Therefore, $\lim _{n \rightarrow+\infty} \frac{a_{n}}{n} \geqslant \frac{2}{3}$.
Similarly, since
$$
\sum_{k=1}^{n} \sqrt{k}<\sum_{k=1}^{n} \frac{2 k}{\sqrt{k-1}+\sqrt{k}}=2 \sum_{k=1}^{n} k(\sqrt{k}-\sqrt{k-1})=2\left((n+1) \sqrt{n}-\sum_{k=1}^{n} \sqrt{k}\right) .
$$

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

which implies $\frac{a_{n}}{n}<\frac{2}{3}\left(1+\frac{1}{n}\right)$. 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 conclusion, $\lim _{n \rightarrow+\infty} \frac{a_{n}}{n}=\frac{2}{3}$.