13. 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 $n \in \mathbf{N}_{+}$,

$\sum_{k=1}^{n} a_{k}^{2}<\frac{7}{6}$.
$$
\begin{array}{l}
\text { (1) } \frac{1}{a_{n+1}}=\frac{n-a_{n}}{(n-1) a_{n}}=\frac{n}{(n-1) a_{n}}-\frac{1}{n-1} \Rightarrow \frac{1}{n a_{n+1}}=\frac{1}{(n-1) a_{n}}-\frac{1}{(n-1) n} \\
\Rightarrow \frac{1}{n a_{n+1}}-\frac{1}{n}=\frac{1}{(n-1) a_{n}}-\frac{1}{n-1} \Rightarrow \frac{1}{(n-1) a_{n}}-\frac{1}{n-1}=\frac{1}{a_{2}}-1=3 \\
\Rightarrow \frac{1}{(n-1) a_{n}}=\frac{1}{n-1}+3=\frac{3 n-2}{n-1} \Rightarrow a_{n}=\frac{1}{3 n-2}\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
(2) $a_{n}^{2}=\frac{1}{(3 n-2)^{2}}<\frac{1}{(3 n-2)^{2}-1}=\frac{1}{(3 n-3)(3 n-1)}=\frac{1}{2}\left(\frac{1}{3 n-3}-\frac{1}{3 n-1}\right)$,

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