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

See explanation.

【Explanation】【Detailed】
$$
\frac{1}{a_{n+1}}=\frac{n-a_{n}}{(n-1) a_{n}}=\frac{n}{(n-1) a_{n}}-\frac{1}{n-1}
$$
(1) From the given, for $n \geq 2$, we have $a_{n+1}$ (n-1) $a_{n}(n-1) a_{n} n-1$. Dividing both sides by $n$ and rearranging, we get $\frac{1}{n a_{n+1}}-\frac{1}{(n-1) a_{n}}=-\left(\frac{1}{n-1}-\frac{1}{n}\right)$
$$
\sum_{\text {then } k=2}^{n-1}\left[\frac{1}{k a_{k+1}}-\frac{1}{(k-1) a}\right]=-\sum_{k=2}^{n-1}\left(\frac{1}{k-1}-\frac{1}{k}\right)=-\left(1-\frac{1}{n-1}\right)
$$

Thus, $\frac{1}{(n-1) a_{n}}-\frac{1}{a_{2}}=-\left(1-\frac{1}{n-1}\right)(n \geq 2)$
Hence, $\frac{1}{(n-1) a_{n}}=\frac{1}{a_{2}}-\left(1-\frac{1}{n-1}\right)=\frac{3 n-2}{n-1}$

Therefore,
$$
a_{n}=\frac{1}{3 n-2}(n \geq 2)
$$

And when $n=1$, it also holds, so
$$
a_{n}=\frac{1}{3 n-2}\left(n \in N_{+}\right)
$$
(2) When $k \geq 2$, we have
$$
a_{k}^{2}=\frac{1}{(3 k-2)^{2}}<\frac{1}{(3 k-4)(3 k-1)}=\frac{1}{3}\left(\frac{1}{3 k-4}-\frac{1}{3 k-1}\right)
$$

Therefore, when $n \geq 2$, we have $\sum_{k=1} a_{k}^{2}=1+\sum_{k=2}^{n} a_{k}^{2}<1+\frac{1}{3} \sum_{k=2}^{n}\left(\frac{1}{3 k-4}-\frac{1}{3 k-1}\right)$
$$
=1+\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3 n-1}\right)<1+\frac{1}{6}=\frac{7}{6}
$$

And when $n=1$,
$$
a_{1}^{2}=1<\frac{7}{6}
$$

Therefore, for all $n \in N_{+}$, we have $\sum_{k=1}^{n} a_{k}^{2}$