11. Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}$ satisfying: $S_{n}+a_{n}=\frac{n-1}{n(n+1)}, n=1,2, \cdots$, then the general term $a_{n}=$ $\qquad$
$$
\begin{array}{l}
\text{When } n \geqslant 2, \text{ we have } S_{n-1}+a_{n-1}=\frac{n-2}{n(n-1)} \Rightarrow 2 a_{n}-a_{n-1}=\frac{n-1}{n(n+1)}-\frac{n-2}{n(n-1)} \\
=\frac{(n-1)^{2}-(n-2)(n+1)}{n(n-1)(n+1)}=\frac{-n+3}{n(n-1)(n+1)}=\frac{(n+1)-2(n-1)}{n(n-1)(n+1)} \\
=\frac{1}{n(n-1)}-\frac{2}{n(n+1)} \Rightarrow 2 a_{n}+\frac{2}{n(n+1)}=a_{n-1}+\frac{1}{n(n-1)}, \text{ and } a_{1}=0,
\end{array}
$$

Therefore, $a_{n}+\frac{1}{n(n+1)}=\left(a_{1}+\frac{1}{2}\right) \cdot \frac{1}{2^{n-1}}=\frac{1}{2^{n}} \Rightarrow a_{n}=\frac{1}{2^{n}}-\frac{1}{n(n+1)}\left(n \in \mathbf{N}^{*}\right)$.