13. Given in the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, when $n \geqslant 2$, the sum of the first $n$ terms $S_{n}$ satisfies $S_{n}^{2}=a_{n} \cdot\left(S_{n}-\frac{1}{2}\right)$.
(1) Find the expression for $S_{n}$;
(2) Let $b_{n}=\frac{S_{n}}{2 n+1}$, the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $T_{n}$, prove: $T_{n}<\frac{1}{2}$.
(1) $S_{n}^{2}=\left(S_{n}-S_{n-1}\right)\left(S_{n}-\frac{1}{2}\right) \Rightarrow S_{n-1}-S_{n}=2 S_{n-1} S_{n} \Rightarrow \frac{1}{S_{n}}-\frac{1}{S_{n-1}}=2$,

and $\frac{1}{S_{1}}=1$, so $\frac{1}{S_{n}}=2 n-1 \Rightarrow S_{n}=\frac{1}{2 n-1}\left(n \in \mathbf{N}_{+}\right)$.
(2) $b_{n}=\frac{1}{(2 n-1)(2 n+1)}=\frac{1}{2}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)$,

thus $T_{n}=\sum_{k=1}^{n} \frac{1}{2}\left(\frac{1}{2 k-1}-\frac{1}{2 k+1}\right)=\frac{1}{2}\left(1-\frac{1}{2 n+1}\right)<\frac{1}{2}$.