11. (20 points) The sequence of real numbers $\left\{a_{n}\right\}$ satisfies
$$
1=a_{0} \leqslant a_{1} \leqslant a_{2} \leqslant \cdots,
$$

The sequence $\left\{b_{n}\right\}$ satisfies
$$
b_{n}=\sum_{k=1}^{n}\left(1-\frac{a_{k-1}}{a_{k}}\right) \frac{1}{\sqrt{a_{k}}}(n=1,2, \cdots) .
$$

Find the largest real number $C$, such that for any real number $x \in[0, C)$, there exists a sequence $\left\{a_{n}\right\}$ satisfying property (1), such that the corresponding sequence $\left\{b_{n}\right\}$ has infinitely many terms greater than $x$.
11. $C_{\max }=2$.

Let $a_{n}=p^{2 n}, p \geqslant 1$ be a constant to be determined. It is easy to prove that $\left\{a_{n}\right\}$ satisfies property (1).
$$
\begin{array}{l}
\text { Then } b_{n}=\left(1-\frac{1}{p^{2}}\right)\left(\frac{1}{p}+\frac{1}{p^{2}}+\cdots+\frac{1}{p^{n}}\right) \\
=\frac{1}{p}+\left(\frac{1}{p}\right)^{2}-\frac{p+1}{p^{n+2}} \text {. }
\end{array}
$$

Since $xx$.
Therefore, for any $n \in \mathbf{Z}_{+}$, and
$$
n>\log _{p} \frac{p+1}{\frac{1}{p}+\left(\frac{1}{p}\right)^{2}-x} \text {, }
$$

we have $b_{n}>x$.
It remains to prove that for any $n \in \mathbf{Z}_{+}$, $b_{n}<2$.
This is because
$$
\begin{array}{l}
b_{n}=\sum_{k=1}^{n} \frac{a_{k}-a_{k-1}}{a_{k}} \cdot \frac{1}{\sqrt{a_{k}}} \\
=\sum_{k=1}^{n} \frac{\left(\sqrt{a_{k}}+\sqrt{a_{k-1}}\right)\left(\sqrt{a_{k}}-\sqrt{a_{k-1}}\right)}{\sqrt{a_{k}} a_{k}} \\
\leqslant 2 \sum_{k=1}^{n} \frac{\sqrt{a_{k}}-\sqrt{a_{k-1}}}{a_{k}} \\
\leqslant 2 \sum_{k=1}^{n} \frac{\sqrt{a_{k}}-\sqrt{a_{k-1}}}{\sqrt{a_{k}} \sqrt{a_{k-1}}} \\
=2 \sum_{k=1}^{n}\left(\frac{1}{\sqrt{a_{k-1}}}-\frac{1}{\sqrt{a_{k}}}\right) \\
=2\left(\frac{1}{\sqrt{a_{0}}}-\frac{1}{\sqrt{a_{n}}}\right)<2 .
\end{array}
$$

In summary, $C_{\max }=2$.