11. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$a_{1}=\sqrt{2}+1$,
$\frac{a_{n}-\sqrt{n}}{a_{n-1}-\sqrt{n-1}}=\sqrt{1+\frac{1}{n}}(n=2,3, \cdots)$.
Let $b_{n}=\frac{1}{a_{n}}$, and let the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ be $S_{n}$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) If for any $n \in \mathbf{Z}_{+}$, we have
$$
\sqrt{n+2} \leqslant \lambda\left(S_{n}+1\right) \leqslant n+101
$$

always holds, find the range of the real number $\lambda$.
(1) By mathematical induction, we have
$$
a_{n}=\sqrt{n+1}+\sqrt{n} \text {. }
$$
$$
\begin{array}{l}
\text { (2) By } b_{n}=\sqrt{n+1}-\sqrt{n} \\
\Rightarrow S_{n}=\sqrt{n+1}-1 \text {. }
\end{array}
$$

From $\sqrt{n+2} \leqslant \lambda\left(S_{n}+1\right) \leqslant n+101$
$$
\Rightarrow \sqrt{\frac{n+2}{n+1}} \leqslant \lambda \leqslant \frac{100}{\sqrt{n+1}}+\sqrt{n+1} \text {. }
$$

Since the above inequality holds for any positive integer, we have
$$
\begin{array}{l}
\frac{\sqrt{6}}{2}=\left(\sqrt{\frac{n+2}{n+1}}\right)_{\max } \\
\leqslant \lambda \leqslant\left(\frac{100}{\sqrt{n+1}}+\sqrt{n+1}\right)_{\text {min }}=20,
\end{array}
$$

i.e., $\frac{\sqrt{6}}{2} \leqslant \lambda \leqslant 20$.