11. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{2}=\frac{1}{4}$, and $a_{n+1}=\frac{(n-1) a_{n}}{n-a_{n}}(n=2,3,4, \cdots)$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove that for all $n \in \mathbf{N}^{*}$, $\sum_{k=1}^{n} a_{k}^{2}<\frac{7}{6}$.
(1) $n a_{n+1}-a_{n} a_{n+1}=(n-1) a_{n} \Rightarrow n a_{n+1}-(n-1) a_{n}=a_{n} a_{n+1}$
$$
\Rightarrow \frac{n}{a_{n}}-\frac{n-1}{a_{n+1}}=1 \Rightarrow \frac{1}{(n-1) a_{n}}-\frac{1}{n a_{n+1}}=\frac{1}{n-1}-\frac{1}{n} \text {, }
$$

then $\frac{1}{a_{2}}-\frac{1}{(n-1) a_{n}}=1-\frac{1}{n-1} \Rightarrow \frac{1}{(n-1) a_{n}}=3+\frac{1}{n-1}=\frac{3 n-2}{n-1} \Rightarrow a_{n}=\frac{1}{3 n-2}(n \geqslant 2)$, when $n=1$ it also satisfies the above formula, so $a_{n}=\frac{1}{3 n-2}\left(n \in \mathbf{N}^{*}\right)$.
(2) When $n \geqslant 2$, $a_{n}^{2}=\frac{1}{(3 n-2)^{2}}<\frac{1}{(3 n-4)(3 n-1)}=\frac{1}{3}\left(\frac{1}{3 n-4}-\frac{1}{3 n-1}\right)$,

thus $\sum_{k=1}^{n} a_{k}^{2}<1+\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3 n-1}\right)=\frac{7}{6}-\frac{1}{3(3 n-1)}<\frac{7}{6}$.
When $n=1$, it is obviously true, so for all $n \in \mathbf{N}^{*}$, we have $\sum_{k=1}^{n} a_{k}^{2}<\frac{7}{6}$.